Rs Aggarwal 2019 for Class 10 Math Chapter 4 - Quadratic Equations

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Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 4 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Math Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 175:

Question 1:

Which of the following are quadratic equation in x ?
(i) $x^{2} - x + 3 = 0$
(ii) $2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0$
(iii) $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$
(iv) $\frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0$
(v) $x^{2} - 3 x - \sqrt{x} + 4 = 0$
(vi) $x - \frac{6}{x} = 3$
(vii) $x + \frac{2}{x} = x^{2}$
(viii) $x^{2} - \frac{1}{x^{2}} = 5$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$

Answer:

$i) (x^{2} - x + 3) is a quadratic polynomial . ∴ x^{2} - x + 3 = 0 is a quadratic equation . ii) C learly, (2 x^{2} + \frac{5}{2} x - \sqrt{3}) is a quadratic polynomial . ∴ 2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0 is a quadratic equation . iii) C learly, (\sqrt{2} x^{2} + 7 x + 5 \sqrt{2}) is a quadratic polynomial . ∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 is a quadratic equation . iv) C learly, (\frac{1}{3} x^{2} + \frac{1}{5} x - 2) is a quadratic polynomial . ∴ \frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0 is a quadratic equation . v) (x^{2} - 3 x - \sqrt{x} + 4) contains a term with \sqrt{x}, i .e, x^{\frac{1}{2}}, where \frac{1}{2} is not a integer . Therefore, it is not a quadratic polynomial . ∴ x^{2} - 3 x - \sqrt{x} + 4 = 0 is not a quadratic equation . vi) x - \frac{6}{x} = 3 \Rightarrow x^{2} - 6 = 3 x \Rightarrow x^{2} - 3 x - 6 = 0 \begin{array}{l} (x^{2} - 3 x - 6) is a quadratic polynomial; therefore, the given \\ equation is quadratic . \end{array} vii) x + \frac{2}{x} = x^{2} \Rightarrow x^{2} + 2 = x^{3} \Rightarrow x^{3} - x^{2} - 2 = 0 (x^{3} - x^{2} - 2) is not a quadratic polynomial . ∴ x^{3} - x^{2} - 2 = 0 is not a quadratic equation . viii) x^{2} - \frac{1}{x^{2}} = 5 \Rightarrow x^{4} - 1 = 5 x^{2} \Rightarrow x^{4} - 5 x^{2} - 1 = 0 (x^{4} - 5 x^{2} - 1) is a polynomial with degree 4 . ∴ x^{4} - 5 x^{2} - 1 = 0 is not a quadratic equation .$
(ix) ${(x + 2)}^{3} = x^{3} - 8$
$\Rightarrow x^{3} + 6 x^{2} + 12 x + 8 = x^{3} - 8 \Rightarrow 6 x^{2} + 12 x + 16 = 0$
This is of the form ax²+ bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)$
$\Rightarrow 6 x^{2} + 4 x + 9 x + 6 = 6 (x^{2} - 3 x + 2) \Rightarrow 6 x^{2} + 13 x + 6 = 6 x^{2} - 18 x + 12 \Rightarrow 31 x - 6 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3$
$\Rightarrow {(\frac{x^{2} + 1}{x})}^{2} = 2 (\frac{x^{2} + 1}{x}) + 3 \Rightarrow {(x^{2} + 1)}^{2} = 2 x (x^{2} + 1) + 3 x^{2} \Rightarrow x^{4} + 2 x^{2} + 1 = 2 x^{3} + 2 x + 3 x^{2} \Rightarrow x^{4} - 2 x^{3} - x^{2} - 2 x + 1 = 0$
This is not of the form ax²+ bx + c = 0.
Hence, the given equation is not a quadratic equation.

Page No 175:

Question 2:

Which of the following are the roots of $3 x^{2} + 2 x - 1 = 0$ ?
(i) −1
(ii) $\frac{1}{3}$
(iii) $- \frac{1}{2}$

Answer:

${The given equation is (3x}^{2} + 2 x - 1 = 0) . (i) x = (- 1) L .H .S . = x^{2} + 2 x - 1 = 3 \times {(- 1)}^{2} + 2 \times (- 1) - 1 = 3 - 2 - 1 = 0 = R .H .S . Thus, (- 1) is a root of {(3x}^{2} + 2 x - 1 = 0) . (ii) On substituting x = \frac{1}{3} in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(\frac{1}{3})}^{2} + 2 \times \frac{1}{3} - 1 = 3^{1} \times \frac{1}{9_{3}} + \frac{2}{3} - 1 = \frac{1 + 2 - 3}{3} = \frac{0}{3} = 0 = R .H .S . Thus, (\frac{1}{3}) is a root of {(3x}^{2} + 2 x - 1 = 0) . (iii) On substituting x = (- \frac{1}{2}) in the given equation, we get : L .H .S . = 3 x^{2} + 2 x - 1 = 3 \times {(- \frac{1}{2})}^{2} + 2^{1} \times (- \frac{1}{2_{1}}) - 1 = 3 \times \frac{1}{4} - 1 - 1 = \frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{- 5}{4} \neq 0 Thus, L .H .S . = R .H .S . Hence, (- \frac{1}{2}) is a not solution of (3 x^{2} + 2 x - 1 = 0) .$

Page No 175:

Question 3:

(i) Find the value of k for which x = 1 is a root of the equation $x^{2} + k x + 3 = 0$ . Also, find the other root.
(ii) Find the values of a and b for which $x = \frac{3}{4} and x = - 2$ are the roots of the equation $a x^{2} + b x - 6 = 0 .$

Answer:

(i)
$It is given that (x = 1) is a root of (x^{2} + k x + 3 = 0) . Therefore, (x = 1) must satisfy the equation . \Rightarrow {(1)}^{2} + k \times 1 + 3 = 0 \Rightarrow k + 4 = 0 \Rightarrow k = - 4 Hence, the required value of k is - 4 .$
So, the equation becomes $x^{2} - 4 x + 3 = 0$
On factorising we get;
$x^{2} - x - 3 x + 3 = 0 x (x - 1) - 3 (x - 1) = 0 (x - 1) (x - 3) = 0 \Rightarrow x - 1 = 0 or x - 3 = 0 \Rightarrow x = 1 or x = 3$
Hence, the other root is 3.

(ii)
$It is given that \frac{3}{4} is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(\frac{3}{4})}^{2} + b \times \frac{3}{4} - 6 = 0 \Rightarrow \frac{9 a}{16} + \frac{3 b}{4} = 6 \Rightarrow \frac{9 a + 12 b}{16} = 6 \Rightarrow 9 a + 12 b - 96 = 0 \Rightarrow 3 a + 4 b = 32 ... (i) Again, (- 2) is a root of a x^{2} + b x - 6 = 0; therefore, we have: a \times {(- 2)}^{2} + b \times (- 2) - 6 = 0 \Rightarrow 4 a - 2 b = 6 \Rightarrow 2 a - b = 3 ... (ii) On multiplying (ii) by 4 and adding the result with (i), w e get: \Rightarrow 3 a + 4 b + 8 a - 4 b = 32 + 12 \Rightarrow 11 a = 44 \Rightarrow a = 4 Putting the value of a in (ii), we get : 2 \times 4 - b = 3 \Rightarrow 8 - b = 3 \Rightarrow b = 5 Hence, the required values of a and b are 4 and 5, respectively.$

Page No 175:

Question 4:

Show that $x = - \frac{b c}{a d}$ is a solution of the quadratic equation $a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0$ .

Answer:

LHS;
Consider the quadratic equation;
$a d^{2} (\frac{a x}{b} + \frac{2 c}{d}) x + b c^{2} = 0$
Put $x = - \frac{b c}{a d}$ in the given equation.
$a d^{2} [\frac{a (- \frac{b c}{a d})}{b} + \frac{2 c}{d}] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [a d (- \frac{b c}{a d}) + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} [- b c + 2 b c] (- \frac{b c}{a d}) + b c^{2} = \frac{a d^{2}}{b d} (b c) (- \frac{b c}{a d}) + b c^{2} = - \frac{a b^{2} c^{2} d^{2}}{a b d^{2}} + b c^{2} = - b c^{2} + b c^{2} = 0 = RHS$
Hence, $x = - \frac{b c}{a d}$ is a solution to the given quadratic equation.

Page No 175:

Question 5:

Solve each of the following quadratic equations:

(2x − 3)(3x + 1) = 0

Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

⇒ x = $\frac{3}{2}$ or x = $- \frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $- \frac{1}{3}$ .

Page No 175:

Question 6:

Solve each of the following quadratic equations:

4x² + 5x = 0

Answer:

4x² + 5x = 0

⇒ x(4x + 5) = 0

⇒ x = 0 or 4x + 5 = 0

⇒ x = 0 or x = $- \frac{5}{4}$

Hence, the roots of the given equation are 0 and $- \frac{5}{4}$ .

Page No 175:

Question 7:

$3 x^{2} - 243 = 0$

Answer:

$Given: 3 x^{2} - 243 = 0 \Rightarrow 3 (x^{2} - 81) = 0 \Rightarrow {(x)}^{2} - {(9)}^{2} = 0 \Rightarrow (x + 9) (x - 9) = 0 \Rightarrow x + 9 = 0 or x - 9 = 0 \Rightarrow x = - 9 or x = 9 Hence, - 9 and 9 are the roots of the equation 3 x^{2} - 243 = 0.$

Page No 175:

Question 8:

Solve each of the following quadratic equations:

$2 x^{2} + x - 6 = 0$

Answer:

We write, $x = 4 x - 3 x$ as $2 x^{2} \times (- 6) = - 12 x^{2} = 4 x \times (- 3 x)$

$∴ 2 x^{2} + x - 6 = 0 \Rightarrow 2 x^{2} + 4 x - 3 x - 6 = 0 \Rightarrow 2 x (x + 2) - 3 (x + 2) = 0 \Rightarrow (x + 2) (2 x - 3) = 0$

$\Rightarrow x + 2 = 0 or 2 x - 3 = 0 \Rightarrow x = - 2 or x = \frac{3}{2}$
Hence, the roots of the given equation are $- 2$ and $\frac{3}{2}$ .

Page No 175:

Question 9:

Solve each of the following quadratic equations:

$x^{2} + 6 x + 5 = 0$

Answer:

We write, $6 x = x + 5 x$ as $x^{2} \times 5 = 5 x^{2} = x \times 5 x$

$∴ x^{2} + 6 x + 5 = 0 \Rightarrow x^{2} + x + 5 x + 5 = 0 \Rightarrow x (x + 1) + 5 (x + 1) = 0 \Rightarrow (x + 1) (x + 5) = 0$

$\Rightarrow x + 1 = 0 or x + 5 = 0 \Rightarrow x = - 1 or x = - 5$

Hence, the roots of the given equation are −1 and −5.

Page No 175:

Question 10:

Solve each of the following quadratic equations:

$9 x^{2} - 3 x - 2 = 0$

Answer:

We write, $- 3 x = 3 x - 6 x$ as $9 x^{2} \times (- 2) = - 18 x^{2} = 3 x \times (- 6 x)$

$∴ 9 x^{2} - 3 x - 2 = 0 \Rightarrow 9 x^{2} + 3 x - 6 x - 2 = 0 \Rightarrow 3 x (3 x + 1) - 2 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x - 2) = 0$
$\Rightarrow 3 x + 1 = 0 or 3 x - 2 = 0 \Rightarrow x = - \frac{1}{3} or x = \frac{2}{3}$
Hence, the roots of the given equation are $- \frac{1}{3}$ and $\frac{2}{3}$ .

Page No 175:

Question 11:

$x^{2} + 12 x + 35 = 0$

Answer:

$Given : x^{2} + 12 x + 35 = 0 \Rightarrow x^{2} + 7 x + 5 x + 35 = 0 \Rightarrow x (x + 7) + 5 (x + 7) = 0 \Rightarrow (x + 5) (x + 7) = 0 \Rightarrow x + 5 = 0 or x + 7 = 0 \Rightarrow x = - 5 or x = - 7 Hence, - 5 and - 7 are the roots of the equation x^{2} + 12 x + 35 = 0.$

Page No 175:

Question 12:

$x^{2} = 18 x - 77$

Answer:

$Given : x^{2} = 18 x - 77 \Rightarrow x^{2} - 18 x + 77 = 0 \Rightarrow x^{2} - (11 x + 7 x) + 77 = 0 \Rightarrow x^{2} - 11 x - 7 x + 77 = 0 \Rightarrow x (x - 11) - 7 (x - 11) = 0 \Rightarrow (x - 7) (x - 11) = 0 \Rightarrow x - 7 = 0 or x - 11 = 0 \Rightarrow x = 7 or x = 11 Hence, 7 and 11 are the roots of the equation x^{2} = 18 x - 77.$

Page No 176:

Question 13:

$6 x^{2} + 11 x + 3 = 0$

Answer:

$Given : 6 x^{2} + 11 x + 3 = 0 \Rightarrow 6 x^{2} + 9 x + 2 x + 3 = 0 \Rightarrow 3 x (2 x + 3) + 1 (2 x + 3) = 0 \Rightarrow (3 x + 1) (2 x + 3) = 0 \Rightarrow 3 x + 1 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 3}{2} Hence, \frac{- 1}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + 11 x + 3 = 0.$

Page No 176:

Question 14:

$6 x^{2} + x - 12 = 0$

Answer:

$Given: 6 x^{2} + x - 12 = 0 \Rightarrow 6 x^{2} + 9 x - 8 x - 12 = 0 \Rightarrow 3 x (2 x + 3) - 4 (2 x + 3) = 0 \Rightarrow (3 x - 4) (2 x + 3) = 0 \Rightarrow 3 x - 4 = 0 or 2 x + 3 = 0 \Rightarrow x = \frac{4}{3} or x = \frac{- 3}{2} Hence, \frac{4}{3} and \frac{- 3}{2} are the roots of the equation 6 x^{2} + x - 12 = 0.$

Page No 176:

Question 15:

Solve each of the following quadratic equations:

$3 x^{2} - 2 x - 1 = 0$

Answer:

We write, $- 2 x = - 3 x + x$ as $3 x^{2} \times (- 1) = - 3 x^{2} = (- 3 x) \times x$

$∴ 3 x^{2} - 2 x - 1 = 0 \Rightarrow 3 x^{2} - 3 x + x - 1 = 0 \Rightarrow 3 x (x - 1) + 1 (x - 1) = 0 \Rightarrow (x - 1) (3 x + 1) = 0$
$\Rightarrow x - 1 = 0 or 3 x + 1 = 0 \Rightarrow x = 1 or x = - \frac{1}{3}$
Hence, the roots of the given equation are $1$ and $- \frac{1}{3}$ .

Page No 176:

Question 16:

$4 x^{2} - 9 x = 100$

Answer:

$Given: 4 x^{2} - 9 x = 100 \Rightarrow 4 x^{2} - 9 x - 100 = 0 \Rightarrow 4 x^{2} - (25 x - 16 x) - 100 = 0 \Rightarrow 4 x^{2} - 25 x + 16 x - 100 = 0 \Rightarrow x (4 x - 25) + 4 (4 x - 25) = 0 \Rightarrow (4 x - 25) (x + 4) = 0 \Rightarrow 4 x - 25 = 0 or x + 4 = 0 \Rightarrow x = \frac{25}{4} or x = - 4 Hence, the roots of the equation are \frac{25}{4} and - 4 .$

Page No 176:

Question 17:

$15 x^{2} - 28 = x$

Answer:

$Given: 15 x^{2} - 28 = x \Rightarrow 15 x^{2} - x - 28 = 0 \Rightarrow 15 x^{2} - (21 x - 20 x) - 28 = 0 \Rightarrow 15 x^{2} - 21 x + 20 x - 28 = 0 \Rightarrow 3 x (5 x - 7) + 4 (5 x - 7) = 0 \Rightarrow (3 x + 4) (5 x - 7) = 0 \Rightarrow 3 x + 4 = 0 or 5 x - 7 = 0 \Rightarrow x = \frac{- 4}{3} or x = \frac{7}{5} H ence, the roots of the equation are \frac{- 4}{3} and \frac{7}{5} .$

Page No 176:

Question 18:

$4 - 11 x = 3 x^{2}$

Answer:

$Given: 4 - 11 x = 3 x^{2} \Rightarrow 3 x^{2} + 11 x - 4 = 0 \Rightarrow 3 x^{2} + 12 x - x - 4 = 0 \Rightarrow 3 x (x + 4) - 1 (x + 4) = 0 \Rightarrow (x + 4) (3 x - 1) = 0 \Rightarrow x + 4 = 0 or 3 x - 1 = 0 \Rightarrow x = - 4 or x = \frac{1}{3} H ence, the roots of the equation are - 4 and \frac{1}{3} .$

Page No 176:

Question 19:

$48 x^{2} - 13 x - 1 = 0$

Answer:

$Given: 48 x^{2} - 13 x - 1 = 0 \Rightarrow 48 x^{2} - (16 x - 3 x) - 1 = 0 \Rightarrow 48 x^{2} - 16 x + 3 x - 1 = 0 \Rightarrow 16 x (3 x - 1) + 1 (3 x - 1) = 0 \Rightarrow (16 x + 1) (3 x - 1) = 0 \Rightarrow 16 x + 1 = 0 or 3 x - 1 = 0 \Rightarrow x = \frac{- 1}{16} or x = \frac{1}{3} Hence, the roots of the equation are \frac{- 1}{16} and \frac{1}{3} .$

Page No 176:

Question 20:

Solve each of the following quadratic equations:

$x^{2} + 2 \sqrt{2} x - 6 = 0$

Answer:

We write, $2 \sqrt{2} x = 3 \sqrt{2} x - \sqrt{2} x$ as $x^{2} \times (- 6) = - 6 x^{2} = 3 \sqrt{2} x \times (- \sqrt{2} x)$

$∴ x^{2} + 2 \sqrt{2} x - 6 = 0 \Rightarrow x^{2} + 3 \sqrt{2} x - \sqrt{2} x - 6 = 0 \Rightarrow x (x + 3 \sqrt{2}) - \sqrt{2} (x + 3 \sqrt{2}) = 0 \Rightarrow (x + 3 \sqrt{2}) (x - \sqrt{2}) = 0$
$\Rightarrow x + 3 \sqrt{2} = 0 or x - \sqrt{2} = 0 \Rightarrow x = - 3 \sqrt{2} or x = \sqrt{2}$
Hence, the roots of the given equation are $- 3 \sqrt{2}$ and $\sqrt{2}$ .

Page No 176:

Question 21:

Solve each of the following quadratic equations:

$\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$

Answer:

Consider $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$
Factorising by splitting the middle term;
$\sqrt{3} x^{2} + 12 x - 2 x - 8 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 4 \sqrt{3}) - 2 (x + 4 \sqrt{3}) = 0 \Rightarrow (\sqrt{3} x - 2) (x + 4 \sqrt{3}) = 0 \Rightarrow \sqrt{3} x - 2 = 0 or x + 4 \sqrt{3} = 0 \Rightarrow x = \frac{2}{\sqrt{3}} or x = - 4 \sqrt{3}$
Hence, the roots of the given equation are $\frac{2}{\sqrt{3}}$ and $- 4 \sqrt{3}$ .

Page No 176:

Question 22:

$\sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0$

Answer:

$Given : \sqrt{3} x^{2} + 11 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} + 9 x + 2 x + 6 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x + 3 \sqrt{3}) + 2 (x + 3 \sqrt{3}) = 0 \Rightarrow (x + 3 \sqrt{3}) (\sqrt{3} x + 2) = 0 \Rightarrow x + 3 \sqrt{3} = 0 or \sqrt{3} x + 2 = 0 \Rightarrow x = - 3 \sqrt{3} or x = \frac{- 2}{\sqrt{3}} = \frac{- 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{- 2 \sqrt{3}}{3} Hence, the roots of the equation are - 3 \sqrt{3} and \frac{- 2 \sqrt{3}}{3} .$

Page No 176:

Question 23:

$3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0$

Answer:

$Given: 3 \sqrt{7} x^{2} + 4 x - \sqrt{7} = 0 \Rightarrow 3 \sqrt{7} x^{2} + 7 x - 3 x - \sqrt{7} = 0 \Rightarrow \sqrt{7} x (3 x + \sqrt{7}) - 1 (3 x + \sqrt{7}) = 0 \Rightarrow (3 x + \sqrt{7}) (\sqrt{7} x - 1) = 0 \Rightarrow 3 x + \sqrt{7} = 0 or \sqrt{7} x - 1 = 0 \Rightarrow x = \frac{- \sqrt{7}}{3} or x = \frac{1}{\sqrt{7}} = \frac{1 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}} = \frac{\sqrt{7}}{7} H ence, the roots of the equation are \frac{- \sqrt{7}}{3} and \frac{\sqrt{7}}{7} .$

Page No 176:

Question 24:

Solve each of the following quadratic equations:

$\sqrt{7} x^{2} - 6 x - 13 \sqrt{7} = 0$

Answer:

We write, $- 6 x = 7 x - 13 x$ as $\sqrt{7} x^{2} \times (- 13 \sqrt{7}) = - 91 x^{2} = 7 x \times (- 13 x)$

$∴ \sqrt{7} x^{2} - 6 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x^{2} + 7 x - 13 x - 13 \sqrt{7} = 0 \Rightarrow \sqrt{7} x (x + \sqrt{7}) - 13 (x + \sqrt{7}) = 0 \Rightarrow (x + \sqrt{7}) (\sqrt{7} x - 13) = 0$
$\Rightarrow x + \sqrt{7} = 0 or \sqrt{7} x - 13 = 0 \Rightarrow x = - \sqrt{7} or x = \frac{13}{\sqrt{7}} = \frac{13 \sqrt{7}}{7}$
Hence, the roots of the given equation are $- \sqrt{7}$ and $\frac{13 \sqrt{7}}{7}$ .

Page No 176:

Question 25:

$4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0$

Answer:

$Given: 4 \sqrt{6} x^{2} - 13 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{6} x^{2} - 16 x + 3 x - 2 \sqrt{6} = 0 \Rightarrow 4 \sqrt{2} x (\sqrt{3} x - 2 \sqrt{2}) + \sqrt{3} (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow (4 \sqrt{2} x + \sqrt{3}) (\sqrt{3} x - 2 \sqrt{2}) = 0 \Rightarrow 4 \sqrt{2} x + \sqrt{3} = 0 or \sqrt{3} x - 2 \sqrt{2} = 0 \Rightarrow x = \frac{- \sqrt{3}}{4 \sqrt{2}} = \frac{- \sqrt{3} \times \sqrt{2}}{4 \sqrt{2} \times \sqrt{2}} = \frac{- \sqrt{6}}{8} or x = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2 \sqrt{6}}{3} H ence, the roots of the equation are \frac{- \sqrt{6}}{8} and \frac{2 \sqrt{6}}{3} .$

Page No 176:

Question 26:

Solve each of the following quadratic equations:

$3 x^{2} - 2 \sqrt{6} x + 2 = 0$ [CBSE 2010, 12]

Answer:

We write, $- 2 \sqrt{6} x = - \sqrt{6} x - \sqrt{6} x$ as $3 x^{2} \times 2 = 6 x^{2} = (- \sqrt{6} x) \times (- \sqrt{6} x)$

$∴ 3 x^{2} - 2 \sqrt{6} x + 2 = 0 \Rightarrow 3 x^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0 \Rightarrow \sqrt{3} x (\sqrt{3} x - \sqrt{2}) - \sqrt{2} (\sqrt{3} x - \sqrt{2}) = 0 \Rightarrow (\sqrt{3} x - \sqrt{2}) (\sqrt{3} x - \sqrt{2}) = 0$
$\Rightarrow {(\sqrt{3} x - \sqrt{2})}^{2} = 0 \Rightarrow \sqrt{3} x - \sqrt{2} = 0 \Rightarrow x = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}$
Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

Page No 176:

Question 27:

Solve each of the following quadratic equations:

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ [CBSE 2011]

Answer:

We write, $- 2 \sqrt{2} x = - 3 \sqrt{2} x + \sqrt{2} x$ as $\sqrt{3} x^{2} \times (- 2 \sqrt{3}) = - 6 x^{2} = (- 3 \sqrt{2} x) \times (\sqrt{2} x)$

$∴ \sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3} = 0 \Rightarrow \sqrt{3} x (x - \sqrt{6}) + \sqrt{2} (x - \sqrt{6}) = 0 \Rightarrow (x - \sqrt{6}) (\sqrt{3} x + \sqrt{2}) = 0$
$\Rightarrow x - \sqrt{6} = 0 or \sqrt{3} x + \sqrt{2} = 0 \Rightarrow x = \sqrt{6} or x = - \frac{\sqrt{2}}{\sqrt{3}} = - \frac{\sqrt{6}}{3}$
Hence, the roots of the given equation are $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ .

Page No 176:

Question 28:

Solve each of the following quadratic equations:

$x^{2} - 3 \sqrt{5} x + 10 = 0$ [CBSE 2011]

Answer:

We write, $- 3 \sqrt{5} x = - 2 \sqrt{5} x - \sqrt{5} x$ as $x^{2} \times 10 = 10 x^{2} = (- 2 \sqrt{5} x) \times (- \sqrt{5} x)$

$∴ x^{2} - 3 \sqrt{5} x + 10 = 0 \Rightarrow x^{2} - 2 \sqrt{5} x - \sqrt{5} x + 10 = 0 \Rightarrow x (x - 2 \sqrt{5}) - \sqrt{5} (x - 2 \sqrt{5}) = 0 \Rightarrow (x - 2 \sqrt{5}) (x - \sqrt{5}) = 0$
$\Rightarrow x - \sqrt{5} = 0 or x - 2 \sqrt{5} = 0 \Rightarrow x = \sqrt{5} or x = 2 \sqrt{5}$
Hence, the roots of the given equation are $\sqrt{5}$ and $2 \sqrt{5}$ .

Page No 176:

Question 29:

Solve each of the following quadratic equations:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ [CBSE 2015]

Answer:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0 \Rightarrow x^{2} - \sqrt{3} x - x + \sqrt{3} = 0 \Rightarrow x (x - \sqrt{3}) - 1 (x - \sqrt{3}) = 0 \Rightarrow (x - \sqrt{3}) (x - 1) = 0$
$\Rightarrow x - \sqrt{3} = 0 or x - 1 = 0 \Rightarrow x = \sqrt{3} or x = 1$
Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

Page No 176:

Question 30:

Solve each of the following quadratic equations:

$x^{2} + 3 \sqrt{3} x - 30 = 0$ [CBSE 2015]

Answer:

We write, $3 \sqrt{3} x = 5 \sqrt{3} x - 2 \sqrt{3} x$ as $x^{2} \times (- 30) = - 30 x^{2} = 5 \sqrt{3} x \times (- 2 \sqrt{3} x)$

$∴ x^{2} + 3 \sqrt{3} x - 30 = 0 \Rightarrow x^{2} + 5 \sqrt{3} x - 2 \sqrt{3} x - 30 = 0 \Rightarrow x (x + 5 \sqrt{3}) - 2 \sqrt{3} (x + 5 \sqrt{3}) = 0 \Rightarrow (x + 5 \sqrt{3}) (x - 2 \sqrt{3}) = 0$
$\Rightarrow x + 5 \sqrt{3} = 0 or x - 2 \sqrt{3} = 0 \Rightarrow x = - 5 \sqrt{3} or x = 2 \sqrt{3}$
Hence, the roots of the given equation are $- 5 \sqrt{3}$ and $2 \sqrt{3}$ .

Page No 176:

Question 31:

Solve each of the following quadratic equations:

$\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ [CBSE 2013]

Answer:

We write, $7 x = 5 x + 2 x$ as $\sqrt{2} x^{2} \times 5 \sqrt{2} = 10 x^{2} = 5 x \times 2 x$

$∴ \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 \Rightarrow \sqrt{2} x^{2} + 5 x + 2 x + 5 \sqrt{2} = 0 \Rightarrow x (\sqrt{2} x + 5) + \sqrt{2} (\sqrt{2} x + 5) = 0 \Rightarrow (\sqrt{2} x + 5) (x + \sqrt{2}) = 0$
$\Rightarrow x + \sqrt{2} = 0 or \sqrt{2} x + 5 = 0 \Rightarrow x = - \sqrt{2} or x = - \frac{5}{\sqrt{2}} = - \frac{5 \sqrt{2}}{2}$
Hence, the roots of the given equation are $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ .

Page No 176:

Question 32:

Solve each of the following quadratic equations:

$5 x^{2} + 13 x + 8 = 0$ [CBSE 2013C]

Answer:

We write, 13x = 5x + 8x as $5 x^{2} \times 8 = 40 x^{2} = 5 x \times 8 x$
$∴ 5 x^{2} + 13 x + 8 = 0 \Rightarrow 5 x^{2} + 5 x + 8 x + 8 = 0 \Rightarrow 5 x (x + 1) + 8 (x + 1) = 0 \Rightarrow (x + 1) (5 x + 8) = 0$
$\Rightarrow x + 1 = 0 or 5 x + 8 = 0 \Rightarrow x = - 1 or x = - \frac{8}{5}$
Hence, $- 1$ and $- \frac{8}{5}$ are the roots of the given equation.

Page No 176:

Question 33:

Solve each of the following quadratic equations:

$x^{2} - (1 + \sqrt{2}) x + \sqrt{2} = 0$

Answer:

$x^{2} - (1 + \sqrt{2}) x + \sqrt{2} = 0 x^{2} - x - \sqrt{2} x + \sqrt{2} = 0 x (x - 1) - \sqrt{2} (x - 1) = 0 (x - \sqrt{2}) (x - 1) = 0 \Rightarrow x - \sqrt{2} = 0 and x - 1 = 0 \Rightarrow x = \sqrt{2} and x = 1$

Page No 176:

Question 34:

$9 x^{2} + 6 x + 1 = 0$

Answer:

$Given : 9 x^{2} + 6 x + 1 = 0 \Rightarrow 9 x^{2} + 3 x + 3 x + 1 = 0 \Rightarrow 3 x (3 x + 1) + 1 (3 x + 1) = 0 \Rightarrow (3 x + 1) (3 x + 1) = 0 \Rightarrow 3 x + 1 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{- 1}{3} or x = \frac{- 1}{3} Hence, \frac{- 1}{3} is the root of the equation 9 x^{2} + 6 x + 1 = 0 .$

Page No 176:

Question 35:

Solve each of the following quadratic equations:

$100 x^{2} - 20 x + 1 = 0$

Answer:

We write, $- 20 x = - 10 x - 10 x$ as $100 x^{2} \times 1 = 100 x^{2} = (- 10 x) \times (- 10 x)$
$∴ 100 x^{2} - 20 x + 1 = 0 \Rightarrow 100 x^{2} - 10 x - 10 x + 1 = 0 \Rightarrow 10 x (10 x - 1) - 1 (10 x - 1) = 0 \Rightarrow (10 x - 1) (10 x - 1) = 0$
$\Rightarrow {(10 x - 1)}^{2} = 0 \Rightarrow 10 x - 1 = 0 \Rightarrow x = \frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

Page No 176:

Question 36:

Solve each of the following quadratic equations:

$2 x^{2} - x + \frac{1}{8} = 0$

Answer:

We write, $- x = - \frac{x}{2} - \frac{x}{2}$ as $2 x^{2} \times \frac{1}{8} = \frac{x^{2}}{4} = (- \frac{x}{2}) \times (- \frac{x}{2})$
$∴ 2 x^{2} - x + \frac{1}{8} = 0 \Rightarrow 2 x^{2} - \frac{x}{2} - \frac{x}{2} + \frac{1}{8} = 0 \Rightarrow 2 x (x - \frac{1}{4}) - \frac{1}{2} (x - \frac{1}{4}) = 0 \Rightarrow (x - \frac{1}{4}) (2 x - \frac{1}{2}) = 0$
$\Rightarrow x - \frac{1}{4} = 0 or 2 x - \frac{1}{2} = 0 \Rightarrow x = \frac{1}{4} or x = \frac{1}{4}$
Hence, $\frac{1}{4}$ is the repeated root of the given equation.

Page No 176:

Question 37:

$10 x - \frac{1}{3} = 3$

Answer:

$Given: 10 x - \frac{1}{x} = 3 \Rightarrow 10 x^{2} - 1 = 3 x [Multiplying both sides by x] \Rightarrow 10 x^{2} - 3 x - 1 = 0 \Rightarrow 10 x^{2} - (5 x - 2 x) - 1 = 0 \Rightarrow 10 x^{2} - 5 x + 2 x - 1 = 0 \Rightarrow 5 x (2 x - 1) + 1 (2 x - 1) = 0 \Rightarrow (2 x - 1) (5 x + 1) = 0 \Rightarrow 2 x - 1 = 0 or 5 x + 1 = 0 \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{5} H ence, the roots of the equation are \frac{1}{2} and \frac{- 1}{5} .$

Page No 176:

Question 38:

$\frac{2}{x^{2}} - \frac{5}{x} + 2 = 0$

Answer:

$Given: \frac{2}{x^{2}} - \frac{5}{x} + 2 = 0 \Rightarrow 2 - 5 x + 2 x^{2} = 0 [Multiplying both side by x^{2}] \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow 2 x^{2} - (4 x + x) + 2 = 0 \Rightarrow 2 x^{2} - 4 x - x + 2 = 0 \Rightarrow 2 x (x - 2) - 1 (x - 2) = 0 \Rightarrow (2 x - 1) (x - 2) = 0 \Rightarrow 2 x - 1 = 0 or x - 2 = 0 \Rightarrow x = \frac{1}{2} or x = 2 H ence, the roots of the equation are \frac{1}{2} and 2 .$

Page No 176:

Question 39:

Solve each of the following quadratic equations:

$2 x^{2} + a x - a^{2} = 0$ [CBSE 2015]

Answer:

We write, $a x = 2 a x - a x$ as $2 x^{2} \times (- a^{2}) = - 2 a^{2} x^{2} = 2 a x \times (- a x)$
$∴ 2 x^{2} + a x - a^{2} = 0 \Rightarrow 2 x^{2} + 2 a x - a x - a^{2} = 0 \Rightarrow 2 x (x + a) - a (x + a) = 0 \Rightarrow (x + a) (2 x - a) = 0$
$\Rightarrow x + a = 0 or 2 x - a = 0 \Rightarrow x = - a or x = \frac{a}{2}$
Hence, $- a$ and $\frac{a}{2}$ are the roots of the given equation.

Page No 176:

Question 40:

Solve each of the following quadratic equations:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ [CBSE 2015]

Answer:

We write, $4 b x = 2 (a + b) x - 2 (a - b) x$ as $4 x^{2} \times [- (a^{2} - b^{2})] = - 4 (a^{2} - b^{2}) x^{2} = 2 (a + b) x \times [- 2 (a - b) x]$
$∴ 4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 2 (a + b) x - 2 (a - b) x - (a - b) (a + b) = 0 \Rightarrow 2 x [2 x + (a + b)] - (a - b) [2 x + (a + b)] = 0 \Rightarrow [2 x + (a + b)] [2 x - (a - b)] = 0$
$\Rightarrow 2 x + (a + b) = 0 or 2 x - (a - b) = 0 \Rightarrow x = - \frac{a + b}{2} or x = \frac{a - b}{2}$
Hence, $- \frac{a + b}{2}$ and $\frac{a - b}{2}$ are the roots of the given equation.

Page No 176:

Question 41:

Solve each of the following quadratic equations:

$4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ [CBSE 2015]

Answer:

We write, $- 4 a^{2} x = - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x$ as $4 x^{2} \times (a^{4} - b^{4}) = 4 (a^{4} - b^{4}) x^{2} = [- 2 (a^{2} + b^{2})] x \times [- 2 (a^{2} - b^{2})] x$
$∴ 4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0 \Rightarrow 4 x^{2} - 2 (a^{2} + b^{2}) x - 2 (a^{2} - b^{2}) x + (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 2 x [2 x - (a^{2} + b^{2})] - (a^{2} - b^{2}) [2 x - (a^{2} + b^{2})] = 0 \Rightarrow [2 x - (a^{2} + b^{2})] [2 x - (a^{2} - b^{2})] = 0$
$\Rightarrow 2 x - (a^{2} + b^{2}) = 0 or 2 x - (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{2} or x = \frac{a^{2} - b^{2}}{2}$
Hence, $\frac{a^{2} + b^{2}}{2}$ and $\frac{a^{2} - b^{2}}{2}$ are the roots of the given equation.

Page No 176:

Question 42:

Solve each of the following quadratic equations:

$x^{2} + 5 x - (a^{2} + a - 6) = 0$ [CBSE 2015]

Answer:

We write, $5 x = (a + 3) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + a - 6)] = - (a^{2} + a - 6) x^{2} = (a + 3) x \times [- (a - 2) x]$
$∴ x^{2} + 5 x - (a^{2} + a - 6) = 0 \Rightarrow x^{2} + (a + 3) x - (a - 2) x - (a + 3) (a - 2) = 0 \Rightarrow x [x + (a + 3)] - (a - 2) [x + (a + 3)] = 0 \Rightarrow [x + (a + 3)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 3) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 3) or x = a - 2$
Hence, $- (a + 3)$ and $(a - 2)$ are the roots of the given equation.

Page No 176:

Question 43:

Solve each of the following quadratic equations:

$x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ [CBSE 2015]

Answer:

We write, $- 2 a x = (2 b - a) x - (2 b + a) x$ as $x^{2} \times [- (4 b^{2} - a^{2})] = - (4 b^{2} - a^{2}) x^{2} = (2 b - a) x \times [- (2 b + a) x]$
$∴ x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0 \Rightarrow x^{2} + (2 b - a) x - (2 b + a) x - (2 b - a) (2 b + a) = 0 \Rightarrow x [x + (2 b - a)] - (2 b + a) [x + (2 b - a)] = 0 \Rightarrow [x + (2 b - a)] [x - (2 b + a)] = 0$
$\Rightarrow x + (2 b - a) = 0 or x - (2 b + a) = 0 \Rightarrow x = - (2 b - a) or x = 2 b + a \Rightarrow x = a - 2 b or x = a + 2 b$
Hence, $a - 2 b$ and $a + 2 b$ are the roots of the given equation.

Page No 176:

Question 44:

Solve each of the following quadratic equations:

$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ [CBSE 2015]

Answer:

We write, $- (2 b - 1) x = - (b - 5) x - (b + 4) x$ as $x^{2} \times (b^{2} - b - 20) = (b^{2} - b - 20) x^{2} = [- (b - 5) x] \times [- (b + 4) x]$
$∴ x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0 \Rightarrow x^{2} - (b - 5) x - (b + 4) x + (b - 5) (b + 4) = 0 \Rightarrow x [x - (b - 5)] - (b + 4) [x - (b - 5)] = 0 \Rightarrow [x - (b - 5)] [x - (b + 4)] = 0$
$\Rightarrow x - (b - 5) = 0 or x - (b + 4) = 0 \Rightarrow x = b - 5 or x = b + 4$
Hence, $b - 5$ and $b + 4$ are the roots of the given equation.

Page No 176:

Question 45:

Solve each of the following quadratic equations:

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ [CBSE 2015]

Answer:

We write, $6 x = (a + 4) x - (a - 2) x$ as $x^{2} \times [- (a^{2} + 2 a - 8)] = - (a^{2} + 2 a - 8) x^{2} = (a + 4) x \times [- (a - 2) x]$
$∴ x^{2} + 6 x - (a^{2} + 2 a - 8) = 0 \Rightarrow x^{2} + (a + 4) x - (a - 2) x - (a + 4) (a - 2) = 0 \Rightarrow x [x + (a + 4)] - (a - 2) [x + (a + 4)] = 0 \Rightarrow [x + (a + 4)] [x - (a - 2)] = 0$
$\Rightarrow x + (a + 4) = 0 or x - (a - 2) = 0 \Rightarrow x = - (a + 4) or x = a - 2$
Hence, $- (a + 4)$ and $(a - 2)$ are the roots of the given equation.

Page No 176:

Question 46:

$a b x^{2} + (b^{2} - a c) x - b c = 0$

Answer:

$Given: a b x^{2} + (b^{2} - a c) x - b c = 0 \Rightarrow a b x^{2} + b^{2} x - a c x - b c = 0 \Rightarrow b x (a x + b) - c (a x + b) = 0 \Rightarrow (b x - c) (a x + b) = 0 \Rightarrow b x - c = 0 or a x + b = 0 \Rightarrow x = \frac{c}{b} or x = \frac{- b}{a} H ence, the roots of the equation are \frac{c}{b} and \frac{- b}{a} .$

Page No 176:

Question 47:

Solve each of the following quadratic equations:

$x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ [CBSE 2012]

Answer:

We write, $- 4 a x = - (b + 2 a) x + (b - 2 a) x$ as $x^{2} \times (- b^{2} + 4 a^{2}) = (- b^{2} + 4 a^{2}) x^{2} = - (b + 2 a) x \times (b - 2 a) x$
$∴ x^{2} - 4 a x - b^{2} + 4 a^{2} = 0 \Rightarrow x^{2} - (b + 2 a) x + (b - 2 a) x - (b - 2 a) (b + 2 a) = 0 \Rightarrow x [x - (b + 2 a)] + (b - 2 a) [x - (b + 2 a)] = 0 \Rightarrow [x - (b + 2 a)] [x + (b - 2 a)] = 0$
$\Rightarrow x - (b + 2 a) = 0 or x + (b - 2 a) = 0 \Rightarrow x = 2 a + b or x = - (b - 2 a) \Rightarrow x = 2 a + b or x = 2 a - b$
Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

Page No 176:

Question 48:

$4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0$

Answer:

$Given : 4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0 \Rightarrow 4 x^{2} - 2 a^{2} x - 2 b^{2} x + a^{2} b^{2} = 0 \Rightarrow 2 x (2 x - a^{2}) - b^{2} (2 x - a^{2}) = 0 \Rightarrow (2 x - b^{2}) (2 x - a^{2}) = 0 \Rightarrow 2 x - b^{2} = 0 or 2 x - a^{2} = 0 \Rightarrow x = \frac{b^{2}}{2} or x = \frac{a^{2}}{2} H ence, the roots of the equation are \frac{b^{2}}{2} and \frac{a^{2}}{2} .$

Page No 176:

Question 49:

$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$

Answer:

$Given : 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 \Rightarrow 12 a b x^{2} - 9 a^{2} x + 8 b^{2} x - 6 a b = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (3 a x + 2 b) (4 b x - 3 a) = 0 \Rightarrow 3 a x + 2 b = 0 or 4 b x - 3 a = 0 \Rightarrow x = \frac{- 2 b}{3 a} or x = \frac{3 a}{4 b} H ence, the roots of the equation are \frac{- 2 b}{3 a} and \frac{3 a}{4 b} .$

Page No 176:

Question 50:

$a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0$

Answer:

$\begin{array}{l} Given : \\ a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0 \Rightarrow b^{2} x (a^{2} x + 1) - 1 (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) (a^{2} x + 1) = 0 \Rightarrow (b^{2} x - 1) = 0 or (a^{2} x + 1) = 0 \Rightarrow x = \frac{1}{b^{2}} or x = \frac{- 1}{a^{2}} H ence, \frac{1}{b^{2}} and \frac{- 1}{a^{2}} are the roots of the given equation. \end{array}$

Page No 176:

Question 51:

Solve each of the following quadratic equations:

$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$ [CBSE 2009]

Answer:

We write, $- 9 (a + b) x = - 3 (2 a + b) x - 3 (a + 2 b) x$ as $9 x^{2} \times (2 a^{2} + 5 a b + 2 b^{2}) = 9 (2 a^{2} + 5 a b + 2 b^{2}) x^{2} = [- 3 (2 a + b) x] \times [- 3 (a + 2 b) x]$
$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0 \Rightarrow 9 x^{2} - 3 (2 a + b) x - 3 (a + 2 b) x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x [3 x - (2 a + b)] - (a + 2 b) [3 x - (2 a + b)] = 0 \Rightarrow [3 x - (2 a + b)] [3 x - (a + 2 b)] = 0$
$\Rightarrow 3 x - (2 a + b) = 0 or 3 x - (a + 2 b) = 0 \Rightarrow x = \frac{2 a + b}{3} or x = \frac{a + 2 b}{3}$
Hence, $\frac{2 a + b}{3}$ and $\frac{a + 2 b}{3}$ are the roots of the given equation.

Page No 176:

Question 52:

Solve each of the following quadratic equations:

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1$ [CBSE 2014]

Answer:

$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1 \Rightarrow \frac{16}{x} - \frac{15}{x + 1} = 1 \Rightarrow \frac{16 x + 16 - 15 x}{x (x + 1)} = 1 \Rightarrow \frac{x + 16}{x^{2} + x} = 1$
$\Rightarrow x^{2} + x = x + 16 (Cross multiplication) \Rightarrow x^{2} - 16 = 0 \Rightarrow (x + 4) (x - 4) = 0 \Rightarrow x + 4 = 0 or x - 4 = 0$
$\Rightarrow x = - 4 or x = 4$
Hence, −4 and 4 are the roots of the given equation.

Page No 176:

Question 53:

Solve each of the following quadratic equations:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2}$ [CBSE 2014]

Answer:

$\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2} \Rightarrow \frac{4}{x} - \frac{5}{2 x + 3} = 3 \Rightarrow \frac{8 x + 12 - 5 x}{x (2 x + 3)} = 3 \Rightarrow \frac{3 x + 12}{2 x^{2} + 3 x} = 3$
$\Rightarrow \frac{x + 4}{2 x^{2} + 3 x} = 1 \Rightarrow 2 x^{2} + 3 x = x + 4 (Cross multiplication) \Rightarrow 2 x^{2} + 2 x - 4 = 0 \Rightarrow x^{2} + x - 2 = 0$
$\Rightarrow x^{2} + 2 x - x - 2 = 0 \Rightarrow x (x + 2) - 1 (x + 2) = 0 \Rightarrow (x + 2) (x - 1) = 0 \Rightarrow x + 2 = 0 or x - 1 = 0$
$\Rightarrow x = - 2 or x = 1$
Hence, −2 and 1 are the roots of the given equation.

Page No 176:

Question 54:

Solve each of the following quadratic equations:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3}$ [CBSE 2014]

Answer:

$\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3} \Rightarrow \frac{3}{x + 1} - \frac{2}{3 x - 1} = \frac{1}{2} \Rightarrow \frac{9 x - 3 - 2 x - 2}{(x + 1) (3 x - 1)} = \frac{1}{2} \Rightarrow \frac{7 x - 5}{3 x^{2} + 2 x - 1} = \frac{1}{2}$
$\Rightarrow 3 x^{2} + 2 x - 1 = 14 x - 10 (Cross multiplication) \Rightarrow 3 x^{2} - 12 x + 9 = 0 \Rightarrow x^{2} - 4 x + 3 = 0 \Rightarrow x^{2} - 3 x - x + 3 = 0$
$\Rightarrow x (x - 3) - 1 (x - 3) = 0 \Rightarrow (x - 3) (x - 1) = 0 \Rightarrow x - 3 = 0 or x - 1 = 0 \Rightarrow x = 3 or x = 1$
Hence, 1 and 3 are the roots of the given equation.

Page No 177:

Question 55:

Solve each of the following quadratic equations:

(i) $\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5$
(ii) $\frac{1}{2 x - 3} + \frac{1}{x - 5} = 1 \frac{1}{9}, x \neq \frac{3}{2}, 5$ [CBSE 2017]

Answer:

(i)
$\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5 \Rightarrow \frac{x + 5 - x + 1}{(x - 1) (x + 5)} = \frac{6}{7} \Rightarrow \frac{6}{x^{2} + 4 x - 5} = \frac{6}{7} \Rightarrow x^{2} + 4 x - 5 = 7$
$\Rightarrow x^{2} + 4 x - 12 = 0 \Rightarrow x^{2} + 6 x - 2 x - 12 = 0 \Rightarrow x (x + 6) - 2 (x + 6) = 0 \Rightarrow (x + 6) (x - 2) = 0$
$\Rightarrow x + 6 = 0 or x - 2 = 0 \Rightarrow x = - 6 or x = 2$
Hence, −6 and 2 are the roots of the given equation.

(ii)
$\frac{1}{2 x - 3} + \frac{1}{x - 5} = 1 \frac{1}{9} \Rightarrow \frac{1}{2 x - 3} + \frac{1}{x - 5} = \frac{10}{9} \Rightarrow \frac{(x - 5) + (2 x - 3)}{(2 x - 3) (x - 5)} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 3 x - 10 x + 15} = \frac{10}{9} \Rightarrow \frac{3 x - 8}{2 x^{2} - 13 x + 15} = \frac{10}{9} \Rightarrow 27 x - 72 = 20 x^{2} - 130 x + 150 \Rightarrow 20 x^{2} - 157 x + 222 = 0 \Rightarrow 20 x^{2} - 120 x - 37 x + 222 = 0 \Rightarrow 20 x (x - 6) - 37 (x - 6) = 0 \Rightarrow (20 x - 37) (x - 6) = 0 \Rightarrow 20 x - 37 = 0 or x - 6 = 0 \Rightarrow x = \frac{37}{20} or x = 6$

Page No 177:

Question 56:

Solve each of the following quadratic equations:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x}$ [CBSE 2013]

Answer:

$\frac{1}{2 a + b + 2 x} = \frac{1}{2 a} + \frac{1}{b} + \frac{1}{2 x} \Rightarrow \frac{1}{2 a + b + 2 x} - \frac{1}{2 x} = \frac{1}{2 a} + \frac{1}{b} \Rightarrow \frac{2 x - 2 a - b - 2 x}{2 x (2 a + b + 2 x)} = \frac{2 a + b}{2 a b} \Rightarrow \frac{- (2 a + b)}{4 x^{2} + 4 a x + 2 b x} = \frac{2 a + b}{2 a b}$
$\Rightarrow 4 x^{2} + 4 a x + 2 b x = - 2 a b \Rightarrow 4 x^{2} + 4 a x + 2 b x + 2 a b = 0 \Rightarrow 4 x (x + a) + 2 b (x + a) = 0 \Rightarrow (x + a) (4 x + 2 b) = 0$
$\Rightarrow x + a = 0 or 4 x + 2 b = 0 \Rightarrow x = - a or x = - \frac{b}{2}$
Hence, $- a$ and $- \frac{b}{2}$ are the roots of the given equation.

Page No 177:

Question 57:

Solve each of the following quadratic equations:

$\frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4 \frac{1}{4}, x \neq 2, 0$

Answer:

$Given : \frac{x + 3}{x - 2} - \frac{1 - x}{x} = 4 \frac{1}{4} \Rightarrow \frac{(x + 3)}{(x - 2)} - \frac{(1 - x)}{x} = \frac{17}{4} \Rightarrow \frac{x (x + 3) - (1 - x) (x - 2)}{(x - 2) x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x - (x - 2 - x^{2} + 2 x)}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{x^{2} + 3 x + x^{2} - 3 x + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} + 2}{x^{2} - 2 x} = \frac{17}{4} \Rightarrow 8 x^{2} + 8 = 17 x^{2} - 34 x [On cross multiplying] \Rightarrow - 9 x^{2} + 34 x + 8 = 0 \Rightarrow 9 x^{2} - 34 x - 8 = 0 \Rightarrow 9 x^{2} - 36 x + 2 x - 8 = 0 \Rightarrow 9 x (x - 4) + 2 (x - 4) = 0 \Rightarrow (x - 4) (9 x + 2) = 0 \Rightarrow x - 4 = 0 or 9 x + 2 = 0 \Rightarrow x = 4 or x = \frac{- 2}{9} H ence, the roots of the equation are 4 and \frac{- 2}{9} .$

Page No 177:

Question 58:

Solve each of the following quadratic equations:

$\frac{3 x - 4}{7} + \frac{7}{3 x - 4} = \frac{5}{2}, x \neq \frac{4}{3}$ [CBSE 2010]

Answer:

$\frac{3 x - 4}{7} + \frac{7}{3 x - 4} = \frac{5}{2}, x \neq \frac{4}{3} \Rightarrow \frac{{(3 x - 4)}^{2} + 49}{7 (3 x - 4)} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 16 + 49}{21 x - 28} = \frac{5}{2} \Rightarrow \frac{9 x^{2} - 24 x + 65}{21 x - 28} = \frac{5}{2}$
$\Rightarrow 18 x^{2} - 48 x + 130 = 105 x - 140 \Rightarrow 18 x^{2} - 153 x + 270 = 0 \Rightarrow 2 x^{2} - 17 x + 30 = 0 \Rightarrow 2 x^{2} - 12 x - 5 x + 30 = 0$
$\Rightarrow 2 x (x - 6) - 5 (x - 6) = 0 \Rightarrow (x - 6) (2 x - 5) = 0 \Rightarrow x - 6 = 0 or 2 x - 5 = 0 \Rightarrow x = 6 or x = \frac{5}{2}$
Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

Page No 177:

Question 59:

Solve each of the following quadratic equations:

(i) $\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1$
(ii) $\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2, x \neq - \frac{1}{2}, 1$ [CBSE 2017]

Answer:

(i)
$\frac{x}{x - 1} + \frac{x - 1}{x} = 4 \frac{1}{4}, x \neq 0, 1 \Rightarrow \frac{x^{2} + {(x - 1)}^{2}}{x (x - 1)} = \frac{17}{4} \Rightarrow \frac{x^{2} + x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4} \Rightarrow \frac{2 x^{2} - 2 x + 1}{x^{2} - x} = \frac{17}{4}$
$\Rightarrow 8 x^{2} - 8 x + 4 = 17 x^{2} - 17 x \Rightarrow 9 x^{2} - 9 x - 4 = 0 \Rightarrow 9 x^{2} - 12 x + 3 x - 4 = 0 \Rightarrow 3 x (3 x - 4) + 1 (3 x - 4) = 0$
$\Rightarrow (3 x - 4) (3 x + 1) = 0 \Rightarrow 3 x - 4 = 0 or 3 x + 1 = 0 \Rightarrow x = \frac{4}{3} or x = - \frac{1}{3}$
Hence, $\frac{4}{3}$ and $- \frac{1}{3}$ are the roots of the given equation.

(ii)
$\frac{x - 1}{2 x + 1} + \frac{2 x + 1}{x - 1} = 2 \Rightarrow \frac{(x - 1)^{2} + (2 x + 1)^{2}}{(2 x + 1) (x - 1)} = 2 \Rightarrow (x^{2} + 1 - 2 x) + (4 x^{2} + 1 + 4 x) = 2 (2 x + 1) (x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 2 (2 x^{2} - x - 1) \Rightarrow 5 x^{2} + 2 x + 2 = 4 x^{2} - 2 x - 2 \Rightarrow x^{2} + 4 x + 4 = 0 \Rightarrow x^{2} + 2 x + 2 x + 4 = 0 \Rightarrow x (x + 2) + 2 (x + 2) = 0 \Rightarrow (x + 2) (x + 2) = 0 \Rightarrow (x + 2) = 0 or (x + 2) = 0 \Rightarrow x = - 2 or x = - 2 \Rightarrow x = - 2$

Page No 177:

Question 60:

Solve each of the following quadratic equations:

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1$ [CBSE 2014]

Answer:

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \frac{4}{15}, x \neq 0, - 1 \Rightarrow \frac{x^{2} + {(x + 1)}^{2}}{x (x + 1)} = \frac{34}{15} \Rightarrow \frac{x^{2} + x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15} \Rightarrow \frac{2 x^{2} + 2 x + 1}{x^{2} + x} = \frac{34}{15}$
$\Rightarrow 30 x^{2} + 30 x + 15 = 34 x^{2} + 34 x \Rightarrow 4 x^{2} + 4 x - 15 = 0 \Rightarrow 4 x^{2} + 10 x - 6 x - 15 = 0 \Rightarrow 2 x (2 x + 5) - 3 (2 x + 5) = 0$
$\Rightarrow (2 x + 5) (2 x - 3) = 0 \Rightarrow 2 x + 5 = 0 or 2 x - 3 = 0 \Rightarrow x = - \frac{5}{2} or x = \frac{3}{2}$
Hence, $- \frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

Page No 177:

Question 61:

Solve each of the following quadratic equations:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = 3 \frac{1}{3}, x \neq 5, 7$ [CBSE 2014]

Answer:

$\frac{x - 4}{x - 5} + \frac{x - 6}{x - 7} = 3 \frac{1}{3}, x \neq 5, 7 \Rightarrow \frac{(x - 4) (x - 7) + (x - 5) (x - 6)}{(x - 5) (x - 7)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 11 x + 28 + x^{2} - 11 x + 30}{x^{2} - 12 x + 35} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 22 x + 58}{x^{2} - 12 x + 35} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 11 x + 29}{x^{2} - 12 x + 35} = \frac{5}{3} \Rightarrow 3 x^{2} - 33 x + 87 = 5 x^{2} - 60 x + 175 \Rightarrow 2 x^{2} - 27 x + 88 = 0$
$\Rightarrow 2 x^{2} - 16 x - 11 x + 88 = 0 \Rightarrow 2 x (x - 8) - 11 (x - 8) = 0 \Rightarrow (x - 8) (2 x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2 x - 11 = 0 \Rightarrow x = 8 or x = \frac{11}{2}$
Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

Page No 177:

Question 62:

Solve each of the following quadratic equations:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4$ [CBSE 2010]

Answer:

$\frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = 3 \frac{1}{3}, x \neq 2, 4 \Rightarrow \frac{(x - 1) (x - 4) + (x - 2) (x - 3)}{(x - 2) (x - 4)} = \frac{10}{3} \Rightarrow \frac{x^{2} - 5 x + 4 + x^{2} - 5 x + 6}{x^{2} - 6 x + 8} = \frac{10}{3} \Rightarrow \frac{2 x^{2} - 10 x + 10}{x^{2} - 6 x + 8} = \frac{10}{3}$
$\Rightarrow \frac{x^{2} - 5 x + 5}{x^{2} - 6 x + 8} = \frac{5}{3} \Rightarrow 3 x^{2} - 15 x + 15 = 5 x^{2} - 30 x + 40 \Rightarrow 2 x^{2} - 15 x + 25 = 0 \Rightarrow 2 x^{2} - 10 x - 5 x + 25 = 0$
$\Rightarrow 2 x (x - 5) - 5 (x - 5) = 0 \Rightarrow (x - 5) (2 x - 5) = 0 \Rightarrow x - 5 = 0 or 2 x - 5 = 0 \Rightarrow x = 5 or x = \frac{5}{2}$
Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.

Page No 177:

Question 63:

$\frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x}, (x \neq 2, 1)$

Answer:

$Given : \frac{1}{(x - 2)} + \frac{2}{(x - 1)} = \frac{6}{x} \Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 1) (x - 2)} = \frac{6}{x} \Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x} \Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12 [O n cross multiplying] \Rightarrow 3 x^{2} - 13 x + 12 = 0 \Rightarrow 3 x^{2} - (9 + 4) x + 12 = 0 \Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0 \Rightarrow 3 x (x - 3) - 4 (x - 3) = 0 \Rightarrow (3 x - 4) (x - 3) = 0 \Rightarrow 3 x - 4 = 0 or x - 3 = 0 \Rightarrow x = \frac{4}{3} or x = 3 H ence, the roots of the equation are \frac{4}{3} and 3.$

Page No 177:

Question 64:

Solve each of the following quadratic equations:

(i) $\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4$ [CBSE 2013C]
(ii) $\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4}, x \neq - 1, - \frac{1}{5}, - 4$ [CBSE 2017]

Answer:

(i)

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4}, x \neq - 1, - 2, - 4 \Rightarrow \frac{x + 2 + 2 x + 2}{(x + 1) (x + 2)} = \frac{5}{x + 4} \Rightarrow \frac{3 x + 4}{x^{2} + 3 x + 2} = \frac{5}{x + 4} \Rightarrow (3 x + 4) (x + 4) = 5 (x^{2} + 3 x + 2)$
$\Rightarrow 3 x^{2} + 16 x + 16 = 5 x^{2} + 15 x + 10 \Rightarrow 2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0$
$\Rightarrow (x - 2) (2 x + 3) = 0 \Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, 2 and $- \frac{3}{2}$ are the roots of the given equation.

(ii)

$\frac{1}{x + 1} + \frac{3}{5 x + 1} = \frac{5}{x + 4} \frac{(5 x + 1) + (3 x + 3)}{(x + 1) (5 x + 1)} = \frac{5}{x + 4} (x + 4) [(5 x + 1) + (3 x + 3)] = 5 (x + 1) (5 x + 1) (x + 4) [8 x + 4] = 25 x^{2} + 30 x + 5 8 x^{2} + 4 x + 32 x + 16 = 25 x^{2} + 30 x + 5 8 x^{2} + 36 x + 16 = 25 x^{2} + 30 x + 5 17 x^{2} - 6 x - 11 = 0 17 x^{2} - 17 x + 11 x - 11 = 0 17 x (x - 1) + 11 (x - 1) = 0 (17 x + 11) (x - 1) = 0 (17 x + 11) = 0 or (x - 1) = 0 x = - \frac{11}{17} or x = 1$

Page No 177:

Question 65:

Solve each of the following quadratic equations:

$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2}$ [CBSE 2014]

Answer:

$3 (\frac{3 x - 1}{2 x + 3}) - 2 (\frac{2 x + 3}{3 x - 1}) = 5, x \neq \frac{1}{3}, - \frac{3}{2} \Rightarrow \frac{3 {(3 x - 1)}^{2} - 2 {(2 x + 3)}^{2}}{(2 x + 3) (3 x - 1)} = 5 \Rightarrow \frac{3 (9 x^{2} - 6 x + 1) - 2 (4 x^{2} + 12 x + 9)}{6 x^{2} + 7 x - 3} = 5 \Rightarrow \frac{27 x^{2} - 18 x + 3 - 8 x^{2} - 24 x - 18}{6 x^{2} + 7 x - 3} = 5$
$\Rightarrow \frac{19 x^{2} - 42 x - 15}{6 x^{2} + 7 x - 3} = 5 \Rightarrow 19 x^{2} - 42 x - 15 = 30 x^{2} + 35 x - 15 \Rightarrow 11 x^{2} + 77 x = 0 \Rightarrow 11 x (x + 7) = 0$
$\Rightarrow x = 0 or x + 7 = 0 \Rightarrow x = 0 or x = - 7$
Hence, 0 and −7 are the roots of the given equation.

Page No 177:

Question 66:

Solve each of the following quadratic equations:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7}$ [CBSE 2014]

Answer:

$3 (\frac{7 x + 1}{5 x - 3}) - 4 (\frac{5 x - 3}{7 x + 1}) = 11, x \neq \frac{3}{5}, - \frac{1}{7} \Rightarrow \frac{3 {(7 x + 1)}^{2} - 4 {(5 x - 3)}^{2}}{(5 x - 3) (7 x + 1)} = 11 \Rightarrow \frac{3 (49 x^{2} + 14 x + 1) - 4 (25 x^{2} - 30 x + 9)}{35 x^{2} - 16 x - 3} = 11 \Rightarrow \frac{147 x^{2} + 42 x + 3 - 100 x^{2} + 120 x - 36}{35 x^{2} - 16 x - 3} = 11$
$\Rightarrow \frac{47 x^{2} + 162 x - 33}{35 x^{2} - 16 x - 3} = 11 \Rightarrow 47 x^{2} + 162 x - 33 = 385 x^{2} - 176 x - 33 \Rightarrow 338 x^{2} - 338 x = 0 \Rightarrow 338 x (x - 1) = 0$
$\Rightarrow x = 0 or x - 1 = 0 \Rightarrow x = 0 or x = 1$
Hence, 0 and 1 are the roots of the given equation.

Page No 177:

Question 67:

$(\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3, (x \neq \frac{- 1}{2}, \frac{3}{4})$

Answer:

$Given : (\frac{4 x - 3}{2 x + 1}) - 10 (\frac{2 x + 1}{4 x - 3}) = 3 P utting \frac{4 x - 3}{2 x + 1} = y, we get: y - \frac{10}{y} = 3 \Rightarrow \frac{y^{2} - 10}{y} = 3 \Rightarrow y^{2} - 10 = 3 y [On cross multiplying] \Rightarrow y^{2} - 3 y - 10 = 0 \Rightarrow y^{2} - (5 - 2) y - 10 = 0 \Rightarrow y^{2} - 5 y + 2 y - 10 = 0 \Rightarrow y (y - 5) + 2 (y - 5) = 0 \Rightarrow (y - 5) (y + 2) = 0 \Rightarrow y - 5 = 0 or y + 2 = 0 \Rightarrow y = 5 or y = - 2 Case I If y = 5, we get : \frac{4 x - 3}{2 x + 1} = 5 \Rightarrow 4 x - 3 = 5 (2 x + 1) [On cross multiplying] \Rightarrow 4 x - 3 = 10 x + 5 \Rightarrow - 6 x = 8 \Rightarrow - 6 x = 8 \Rightarrow x = - \frac{8^{4}}{6^{3}} \Rightarrow x = - \frac{4}{3} Case II If y = - 2, we get : \frac{4 x - 3}{2 x + 1} = - 2 \begin{array}{l} \Rightarrow 4 x - 3 = - 2 (2 x + 1) \\ \Rightarrow 4 x - 3 = - 4 x - 2 \Rightarrow 8 x = 1 \Rightarrow x = \frac{1}{8} Hence, the roots of the equation are - \frac{4}{3} and \frac{1}{8} . \end{array}$

Page No 177:

Question 68:

${(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0, (x \neq - 1)$

Answer:

$Given : {(\frac{x}{x + 1})}^{2} - 5 (\frac{x}{x + 1}) + 6 = 0 P utting \frac{x}{x + 1} = y, we get : y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - 5 y + 6 = 0 \Rightarrow y^{2} - (3 + 2) y + 6 = 0 \Rightarrow y^{2} - 3 y - 2 y + 6 = 0 \Rightarrow y (y - 3) - 2 (y - 3) = 0 \Rightarrow (y - 3) (y - 2) = 0 \Rightarrow y - 3 = 0 or y - 2 = 0 \Rightarrow y = 3 or y = 2 Case I If y = 3, w e g e t : \frac{x}{x + 1} = 3 \Rightarrow x = 3 (x + 1) [On cross multiplying] \Rightarrow x = 3 x + 3 \Rightarrow x = \frac{- 3}{2} Case II If y = 2, we get : \frac{x}{x + 1} = 2 \Rightarrow x = 2 (x + 1) \Rightarrow x = 2 x + 2 \Rightarrow - x = 2 \Rightarrow x = - 2 Hence, the roots of the equation are \frac{- 3}{2} and - 2 .$

Page No 177:

Question 69:

$\frac{a}{(x - b)} + \frac{b}{(x - a)} = 2, (x \neq b, a)$

Answer:

$\begin{array}{l} \frac{a}{(x - b)} + \frac{b}{(x - a)} = 2 \\ \begin{array}{l} \Rightarrow [\frac{a}{(x - b)} - 1] + [\frac{b}{(x - a)} - 1] = 0 \\ \Rightarrow \frac{a - (x - b)}{x - b} + \frac{b - (x - a)}{x - a} = 0 \end{array} \end{array} \Rightarrow \frac{a - x + b}{x - b} + \frac{a - x + b}{x - a} = 0 \Rightarrow (a - x + b) [\frac{1}{(x - b)} + \frac{1}{(x - a)}] = 0 \Rightarrow (a - x + b) [\frac{(x - a) + (x - b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [\frac{2 x - (a + b)}{(x - b) (x - a)}] = 0 \Rightarrow (a - x + b) [2 x - (a + b)] = 0 \Rightarrow a - x + b = 0 or 2 x - (a + b) = 0 \Rightarrow x = a + b or x = \frac{a + b}{2} Hence, the roots of the equation are (a + b) and (\frac{a + b}{2}) .$

Page No 177:

Question 70:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b), (x \neq \frac{1}{a}, \frac{1}{b})$

Answer:

$\frac{a}{(a x - 1)} + \frac{b}{(b x - 1)} = (a + b) \Rightarrow [\frac{a}{(a x - 1)} - b] + [\frac{b}{(b x - 1)} - a] = 0 \Rightarrow \frac{a - b (a x - 1)}{a x - 1} + \frac{b - a (b x - 1)}{b x - 1} = 0 \Rightarrow \frac{a - a b x + b}{a x - 1} + \frac{a - a b x + b}{b x - 1} = 0 \Rightarrow (a - a b x + b) [\frac{1}{(a x - 1)} + \frac{1}{(b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(b x - 1) + (a x - 1)}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [\frac{(a + b) x - 2}{(a x - 1) (b x - 1)}] = 0 \Rightarrow (a - a b x + b) [(a + b) x - 2] = 0 \Rightarrow a - a b x + b = 0 or (a + b) x - 2 = 0 \Rightarrow x = \frac{(a + b)}{a b} or x = \frac{2}{(a + b)} Hence, the roots of the equation are \frac{(a + b)}{a b} and \frac{2}{(a + b)} .$

Page No 177:

Question 71:

$3^{(x + 2)} + 3^{- x} = 10$

Answer:

$3^{(x + 2)} + 3^{- x} = 10 3^{x} . 9 + \frac{1}{3^{x}} = 10 Let 3^{x} be equal to y . ∴ 9 y + \frac{1}{y} = 10 \Rightarrow 9 y^{2} + 1 = 10 y \Rightarrow 9 y^{2} - 10 y + 1 = 0 \Rightarrow (y - 1) (9 y - 1) = 0 \Rightarrow y - 1 = 0 or 9 y - 1 = 0 \Rightarrow y = 1 or y = \frac{1}{9} \Rightarrow 3^{x} = 1 {or 3}^{x} = \frac{1}{9} \Rightarrow 3^{x} = 3^{0} {or 3}^{x} = {3^{-}}^{2} \Rightarrow x = 0 or x = - 2 Hence, 0 and - 2 are the roots of the given equation .$

Page No 177:

Question 72:

$4^{(x + 1)} + 4^{(1 - x)} = 10$

Answer:

$Given : 4^{(x + 1)} + 4^{(1 - x)} = 10 \Rightarrow 4^{x} .4 + 4^{1} . \frac{1}{4^{x}} = 10 {Let 4}^{x} be y . \begin{array}{l} ∴ 4 y + \frac{4}{y} = 10 \\ \Rightarrow 4 y^{2} - 10 y + 4 = 0 \end{array} \Rightarrow 4 y^{2} - 8 y - 2 y + 4 = 0 \Rightarrow 4 y (y - 2) - 2 (y - 2) = 0 \begin{array}{l} \Rightarrow y = 2 or y = \frac{2}{4} = \frac{1}{2} \\ \Rightarrow 4^{x} = 2 or \frac{1}{2} \end{array} \Rightarrow 4^{x} = 2^{2 x} = 2^{1} or 2^{2 x} = 2^{- 1} \Rightarrow x = \frac{1}{2} or x = \frac{- 1}{2} Hence, \frac{1}{2} and \frac{- 1}{2} are roots of the given equation .$

Page No 177:

Question 73:

$2^{2 x} - 3 . 2^{(x + 2)} + 32 = 0$

Answer:

$Given : 2^{2 x} - {3.2}^{(x + 2)} + 32 = 0 \Rightarrow {(2^{x})}^{2} - {3.2}^{x} {.2}^{2} + 32 = 0 {Let 2}^{x} be y . ∴ y^{2} - 12 y + 32 = 0 \Rightarrow y^{2} - 8 y - 4 y + 32 = 0 \Rightarrow y (y - 8) - 4 (y - 8) = 0 \Rightarrow (y - 8) = 0 or (y - 4) = 0 \Rightarrow y = 8 or y = 4 ∴ 2^{x} = 8 or 2^{x} = 4 \Rightarrow 2^{x} = 2^{3} or 2^{x} = 2^{2} \Rightarrow x = 2 or 3 Hence, 2 and 3 are the roots of the given equation .$

Page No 185:

Question 1:

Solve each of the following equations by using the method of completing the square:

$x^{2} - 6 x + 3 = 0$

Answer:

$x^{2} - 6 x + 3 = 0 \Rightarrow x^{2} - 6 x = - 3 \Rightarrow x^{2} - 2 \times x \times 3 + 3^{2} = - 3 + 3^{2} (Adding 3^{2} on both sides) \Rightarrow {(x - 3)}^{2} = - 3 + 9 = 6$
$\Rightarrow x - 3 = \pm \sqrt{6} (Taking square root on both sides) \Rightarrow x - 3 = \sqrt{6} or x - 3 = - \sqrt{6} \Rightarrow x = 3 + \sqrt{6} or x = 3 - \sqrt{6}$
Hence, $3 + \sqrt{6}$ and $3 - \sqrt{6}$ are the roots of the given equation.

Page No 185:

Question 2:

Solve each of the following equations by using the method of completing the square:

$x^{2} - 4 x + 1 = 0$

Answer:

$x^{2} - 4 x + 1 = 0 \Rightarrow x^{2} - 4 x = - 1 \Rightarrow x^{2} - 2 \times x \times 2 + 2^{2} = - 1 + 2^{2} (Adding 2^{2} on both sides) \Rightarrow {(x - 2)}^{2} = - 1 + 4 = 3$
$\Rightarrow x - 2 = \pm \sqrt{3} (Taking square root on both sides) \Rightarrow x - 2 = \sqrt{3} or x - 2 = - \sqrt{3} \Rightarrow x = 2 + \sqrt{3} or x = 2 - \sqrt{3}$
Hence, $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are the roots of the given equation.

Page No 185:

Question 3:

Solve each of the following equations by using the method of completing the square:

$x^{2} + 8 x - 2 = 0$

Answer:

$x^{2} + 8 x - 2 = 0 \Rightarrow x^{2} + 8 x = 2 \Rightarrow x^{2} + 2 \times x \times 4 + 4^{2} = 2 + 4^{2} (Adding 4^{2} on both sides) \Rightarrow {(x + 4)}^{2} = 2 + 16 = 18$
$\Rightarrow x + 4 = \pm \sqrt{18} = \pm 3 \sqrt{2} (Taking square root on both sides) \Rightarrow x + 4 = 3 \sqrt{2} or x + 4 = - 3 \sqrt{2} \Rightarrow x = - 4 + 3 \sqrt{2} or x = - 4 - 3 \sqrt{2}$
Hence, $(- 4 + 3 \sqrt{2})$ and $(- 4 - 3 \sqrt{2})$ are the roots of the given equation.

Page No 185:

Question 4:

Solve each of the following equations by using the method of completing the square:

$4 x^{2} + 4 \sqrt{3} x + 3 = 0$

Answer:

$4 x^{2} + 4 \sqrt{3} x + 3 = 0 \Rightarrow 4 x^{2} + 4 \sqrt{3} x = - 3 \Rightarrow {(2 x)}^{2} + 2 \times 2 x \times \sqrt{3} + {(\sqrt{3})}^{2} = - 3 + {(\sqrt{3})}^{2} [Adding {(\sqrt{3})}^{2} on both sides] \Rightarrow {(2 x + \sqrt{3})}^{2} = - 3 + 3 = 0$
$\Rightarrow 2 x + \sqrt{3} = 0 \Rightarrow x = - \frac{\sqrt{3}}{2}$
Hence, $- \frac{\sqrt{3}}{2}$ is the repeated root of the given equation.

Page No 185:

Question 5:

Solve each of the following equations by using the method of completing the square:

$2 x^{2} + 5 x - 3 = 0$

Answer:

$2 x^{2} + 5 x - 3 = 0 \Rightarrow 4 x^{2} + 10 x - 6 = 0 (Multiplying both sides by 2) \Rightarrow 4 x^{2} + 10 x = 6 \Rightarrow {(2 x)}^{2} + 2 \times 2 x \times \frac{5}{2} + {(\frac{5}{2})}^{2} = 6 + {(\frac{5}{2})}^{2} [Adding {(\frac{5}{2})}^{2} on both sides]$
$\Rightarrow {(2 x + \frac{5}{2})}^{2} = 6 + \frac{25}{4} = \frac{24 + 25}{4} = \frac{49}{4} = {(\frac{7}{2})}^{2} \Rightarrow 2 x + \frac{5}{2} = \pm \frac{7}{2} (Taking square root on both sides) \Rightarrow 2 x + \frac{5}{2} = \frac{7}{2} or 2 x + \frac{5}{2} = - \frac{7}{2} \Rightarrow 2 x = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1 or 2 x = - \frac{7}{2} - \frac{5}{2} = - \frac{12}{2} = - 6$
$\Rightarrow x = \frac{1}{2} or x = - 3$
Hence, $\frac{1}{2}$ and $- 3$ are the roots of the given equation.

Page No 185:

Question 6:

Solve each of the following equations by using the method of completing the square:

$3 x^{2} - x - 2 = 0$

Answer:

$3 x^{2} - x - 2 = 0 \Rightarrow 9 x^{2} - 3 x - 6 = 0 (Multiplying both sides by 3) \Rightarrow 9 x^{2} - 3 x = 6 \Rightarrow {(3 x)}^{2} - 2 \times 3 x \times \frac{1}{2} + {(\frac{1}{2})}^{2} = 6 + {(\frac{1}{2})}^{2} [Adding {(\frac{1}{2})}^{2} on both sides]$
$\Rightarrow {(3 x - \frac{1}{2})}^{2} = 6 + \frac{1}{4} = \frac{25}{4} = {(\frac{5}{2})}^{2} \Rightarrow 3 x - \frac{1}{2} = \pm \frac{5}{2} (Taking square root on both sides) \Rightarrow 3 x - \frac{1}{2} = \frac{5}{2} or 3 x - \frac{1}{2} = - \frac{5}{2} \Rightarrow 3 x = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3 or 3 x = - \frac{5}{2} + \frac{1}{2} = - \frac{4}{2} = - 2$
$\Rightarrow x = 1 or x = - \frac{2}{3}$
Hence, 1 and $- \frac{2}{3}$ are the roots of the given equation.

Page No 185:

Question 7:

Solve each of the following equations by using the method of completing the square:

$8 x^{2} - 14 x - 15 = 0$

Answer:

$8 x^{2} - 14 x - 15 = 0 \Rightarrow 16 x^{2} - 28 x - 30 = 0 (Multiplying both sides by 2) \Rightarrow 16 x^{2} - 28 x = 30 \Rightarrow {(4 x)}^{2} - 2 \times 4 x \times \frac{7}{2} + {(\frac{7}{2})}^{2} = 30 + {(\frac{7}{2})}^{2} [Adding {(\frac{7}{2})}^{2} on both sides]$
$\Rightarrow {(4 x - \frac{7}{2})}^{2} = 30 + \frac{49}{4} = \frac{169}{4} = {(\frac{13}{2})}^{2} \Rightarrow 4 x - \frac{7}{2} = \pm \frac{13}{2} (Taking square root on both sides) \Rightarrow 4 x - \frac{7}{2} = \frac{13}{2} or 4 x - \frac{7}{2} = - \frac{13}{2} \Rightarrow 4 x = \frac{13}{2} + \frac{7}{2} = \frac{20}{2} = 10 or 4 x = - \frac{13}{2} + \frac{7}{2} = - \frac{6}{2} = - 3$
$\Rightarrow x = \frac{5}{2} or x = - \frac{3}{4}$
Hence, $\frac{5}{2}$ and $- \frac{3}{4}$ are the roots of the given equation.

Page No 185:

Question 8:

Solve each of the following equations by using the method of completing the square:

$7 x^{2} + 3 x - 4 = 0$

Answer:

$7 x^{2} + 3 x - 4 = 0 \Rightarrow 49 x^{2} + 21 x - 28 = 0 (Multiplying both sides by 7) \Rightarrow 49 x^{2} + 21 x = 28 \Rightarrow {(7 x)}^{2} + 2 \times 7 x \times \frac{3}{2} + {(\frac{3}{2})}^{2} = 28 + {(\frac{3}{2})}^{2} [Adding {(\frac{3}{2})}^{2} on both sides]$
$\Rightarrow {(7 x + \frac{3}{2})}^{2} = 28 + \frac{9}{4} = \frac{121}{4} = {(\frac{11}{2})}^{2} \Rightarrow 7 x + \frac{3}{2} = \pm \frac{11}{2} (Taking square root on both sides) \Rightarrow 7 x + \frac{3}{2} = \frac{11}{2} or 7 x + \frac{3}{2} = - \frac{11}{2} \Rightarrow 7 x = \frac{11}{2} - \frac{3}{2} = \frac{8}{2} = 4 or 7 x = - \frac{11}{2} - \frac{3}{2} = - \frac{14}{2} = - 7$
$\Rightarrow x = \frac{4}{7} or x = - 1$
Hence, $\frac{4}{7}$ and −1 are the roots of the given equation.

Page No 185:

Question 9:

Solve each of the following equations by using the method of completing the square:

$3 x^{2} - 2 x - 1 = 0$

Answer:

$3 x^{2} - 2 x - 1 = 0 \Rightarrow 9 x^{2} - 6 x - 3 = 0 (Multiplying both sides by 3) \Rightarrow 9 x^{2} - 6 x = 3 \Rightarrow {(3 x)}^{2} - 2 \times 3 x \times 1 + 1^{2} = 3 + 1^{2} [Adding 1^{2} on both sides]$
$\Rightarrow {(3 x - 1)}^{2} = 3 + 1 = 4 = {(2)}^{2} \Rightarrow 3 x - 1 = \pm 2 (Taking square root on both sides) \Rightarrow 3 x - 1 = 2 or 3 x - 1 = - 2 \Rightarrow 3 x = 3 or 3 x = - 1$
$\Rightarrow x = 1 or x = - \frac{1}{3}$
Hence, 1 and $- \frac{1}{3}$ are the roots of the given equation.

Page No 185:

Question 10:

Solve each of the following equations by using the method of completing the square:

$5 x^{2} - 6 x - 2 = 0$

Answer:

$5 x^{2} - 6 x - 2 = 0 \Rightarrow 25 x^{2} - 30 x - 10 = 0 (Multiplying both sides by 5) \Rightarrow 25 x^{2} - 30 x = 10 \Rightarrow {(5 x)}^{2} - 2 \times 5 x \times 3 + 3^{2} = 10 + 3^{2} (Adding 3^{2} on both sides)$
$\Rightarrow {(5 x - 3)}^{2} = 10 + 9 = 19 \Rightarrow 5 x - 3 = \pm \sqrt{19} (Taking square root on both sides) \Rightarrow 5 x - 3 = \sqrt{19} or 5 x - 3 = - \sqrt{19} \Rightarrow 5 x = 3 + \sqrt{19} or 5 x = 3 - \sqrt{19}$
$\Rightarrow x = \frac{3 + \sqrt{19}}{5} or x = \frac{3 - \sqrt{19}}{5}$
Hence, $\frac{3 + \sqrt{19}}{5}$ and $\frac{3 - \sqrt{19}}{5}$ are the roots of the given equation.

Page No 185:

Question 11:

Solve each of the following equations by using the method of completing the square:

$\frac{2}{x^{2}} - \frac{5}{x} + 2 = 0$

Answer:

$\frac{2}{x^{2}} - \frac{5}{x} + 2 = 0 \Rightarrow \frac{2 - 5 x + 2 x^{2}}{x^{2}} = 0 \Rightarrow 2 x^{2} - 5 x + 2 = 0 \Rightarrow 4 x^{2} - 10 x + 4 = 0 (Multiplying both sides by 2)$
$\Rightarrow 4 x^{2} - 10 x = - 4 \Rightarrow {(2 x)}^{2} - 2 \times 2 x \times \frac{5}{2} + {(\frac{5}{2})}^{2} = - 4 + {(\frac{5}{2})}^{2} [Adding {(\frac{5}{2})}^{2} on both sides] \Rightarrow {(2 x - \frac{5}{2})}^{2} = - 4 + \frac{25}{4} = \frac{9}{4} = {(\frac{3}{2})}^{2} \Rightarrow 2 x - \frac{5}{2} = \pm \frac{3}{2} (Taking square root on both sides)$
$\Rightarrow 2 x - \frac{5}{2} = \frac{3}{2} or 2 x - \frac{5}{2} = - \frac{3}{2} \Rightarrow 2 x = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 or 2 x = - \frac{3}{2} + \frac{5}{2} = \frac{2}{2} = 1 \Rightarrow x = 2 or x = \frac{1}{2}$
Hence, 2 and $\frac{1}{2}$ are the roots of the given equation.

Page No 185:

Question 12:

Solve each of the following equations by using the method of completing the square:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$

Answer:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0 \Rightarrow 4 x^{2} + 4 b x = a^{2} - b^{2} \Rightarrow {(2 x)}^{2} + 2 \times 2 x \times b + b^{2} = a^{2} - b^{2} + b^{2} (Adding b^{2} on both sides) \Rightarrow {(2 x + b)}^{2} = a^{2}$
$\Rightarrow 2 x + b = \pm a (Taking square root on both sides) \Rightarrow 2 x + b = a or 2 x + b = - a \Rightarrow 2 x = a - b or 2 x = - a - b \Rightarrow x = \frac{a - b}{2} or x = - \frac{a + b}{2}$
Hence, $\frac{a - b}{2}$ and $- \frac{a + b}{2}$ are the roots of the given equation.

Page No 185:

Question 13:

Solve each of the following equations by using the method of completing the square:

$x^{2} - (\sqrt{2} + 1) x + \sqrt{2} = 0$

Answer:

$x^{2} - (\sqrt{2} + 1) x + \sqrt{2} = 0 \Rightarrow x^{2} - (\sqrt{2} + 1) x = - \sqrt{2} \Rightarrow x^{2} - 2 \times x \times (\frac{\sqrt{2} + 1}{2}) + {(\frac{\sqrt{2} + 1}{2})}^{2} = - \sqrt{2} + {(\frac{\sqrt{2} + 1}{2})}^{2} [Adding {(\frac{\sqrt{2} + 1}{2})}^{2} on both sides]$
$\Rightarrow {[x - (\frac{\sqrt{2} + 1}{2})]}^{2} = \frac{- 4 \sqrt{2} + 2 + 1 + 2 \sqrt{2}}{4} = \frac{2 - 2 \sqrt{2} + 1}{4} = {(\frac{\sqrt{2} - 1}{2})}^{2} \Rightarrow x - (\frac{\sqrt{2} + 1}{2}) = \pm (\frac{\sqrt{2} - 1}{2}) (Taking square root on both sides) \Rightarrow x - (\frac{\sqrt{2} + 1}{2}) = (\frac{\sqrt{2} - 1}{2}) or x - (\frac{\sqrt{2} + 1}{2}) = - (\frac{\sqrt{2} - 1}{2}) \Rightarrow x = \frac{\sqrt{2} + 1}{2} + \frac{\sqrt{2} - 1}{2} or x = \frac{\sqrt{2} + 1}{2} - \frac{\sqrt{2} - 1}{2}$
$\Rightarrow x = \frac{2 \sqrt{2}}{2} = \sqrt{2} or x = \frac{2}{2} = 1$
Hence, $\sqrt{2}$ and 1 are the roots of the given equation.

Page No 185:

Question 14:

Solve each of the following equations by using the method of completing the square:

$\sqrt{2} x^{2} - 3 x - 2 \sqrt{2} = 0$

Answer:

$\sqrt{2} x^{2} - 3 x - 2 \sqrt{2} = 0 \Rightarrow 2 x^{2} - 3 \sqrt{2} x - 4 = 0 (Multiplying both sides by \sqrt{2}) \Rightarrow 2 x^{2} - 3 \sqrt{2} x = 4 \Rightarrow {(\sqrt{2} x)}^{2} - 2 \times \sqrt{2} x \times \frac{3}{2} + {(\frac{3}{2})}^{2} = 4 + {(\frac{3}{2})}^{2} [Adding {(\frac{3}{2})}^{2} on both sides]$
$\Rightarrow {(\sqrt{2} x - \frac{3}{2})}^{2} = 4 + \frac{9}{4} = \frac{25}{4} = {(\frac{5}{2})}^{2} \Rightarrow \sqrt{2} x - \frac{3}{2} = \pm \frac{5}{2} (Taking square root on both sides) \Rightarrow \sqrt{2} x - \frac{3}{2} = \frac{5}{2} or \sqrt{2} x - \frac{3}{2} = - \frac{5}{2} \Rightarrow \sqrt{2} x = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4 or \sqrt{2} x = - \frac{5}{2} + \frac{3}{2} = - \frac{2}{2} = - 1$
$\Rightarrow x = \frac{4}{\sqrt{2}} = 2 \sqrt{2} or x = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$
Hence, $2 \sqrt{2}$ and $- \frac{\sqrt{2}}{2}$ are the roots of the given equation.

Page No 185:

Question 15:

Solve each of the following equations by using the method of completing the square:

$\sqrt{3} x^{2} + 10 x + 7 \sqrt{3} = 0$

Answer:

$\sqrt{3} x^{2} + 10 x + 7 \sqrt{3} = 0 \Rightarrow 3 x^{2} + 10 \sqrt{3} x + 21 = 0 (Multiplying both sides by \sqrt{3}) \Rightarrow 3 x^{2} + 10 \sqrt{3} x = - 21 \Rightarrow {(\sqrt{3} x)}^{2} + 2 \times \sqrt{3} x \times 5 + 5^{2} = - 21 + 5^{2} (Adding 5^{2} on both sides)$
$\Rightarrow {(\sqrt{3} x + 5)}^{2} = - 21 + 25 = 4 = 2^{2} \Rightarrow \sqrt{3} x + 5 = \pm 2 (Taking square root on both sides) \Rightarrow \sqrt{3} x + 5 = 2 or \sqrt{3} x + 5 = - 2 \Rightarrow \sqrt{3} x = - 3 or \sqrt{3} x = - 7$
$\Rightarrow x = - \frac{3}{\sqrt{3}} = - \sqrt{3} or x = - \frac{7}{\sqrt{3}} = - \frac{7 \sqrt{3}}{3}$
Hence, $- \sqrt{3}$ and $- \frac{7 \sqrt{3}}{3}$ are the roots of the given equation.

Page No 185:

Question 16:

By using the method of completing the square, show that the equation $2 x^{2} + x + 4 = 0$ has no real roots.

Answer:

$2 x^{2} + x + 4 = 0 \Rightarrow 4 x^{2} + 2 x + 8 = 0 (Multiplying both sides by 2) \Rightarrow 4 x^{2} + 2 x = - 8 \Rightarrow {(2 x)}^{2} + 2 \times 2 x \times \frac{1}{2} + {(\frac{1}{2})}^{2} = - 8 + {(\frac{1}{2})}^{2} [Adding {(\frac{1}{2})}^{2} on both sides]$
$\Rightarrow {(2 x + \frac{1}{2})}^{2} = - 8 + \frac{1}{4} = - \frac{31}{4} < 0$
But, ${(2 x + \frac{1}{2})}^{2}$ cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.

Page No 191:

Question 1:

Find the discriminant of each of the following equations:

(i) $2 x^{2} - 7 x + 6 = 0$
(ii) $3 x^{2} - 2 x + 8 = 0$
(iii) $2 x^{2} - 5 \sqrt{2} x + 4 = 0$
(iv) $\sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0$
(v) $(x - 1) (2 x - 1) = 0$
(vi) $1 - x = 2 x^{2}$ 2x2−7x+6=02x2−7x+6=0

Answer:

(i) 2 x^{2} - 7 x + 6 = 0 Here, a = 2, b = - 7, c = 6 Discriminant D is diven by : D = b^{2} - 4 a c = {(- 7)}^{2} - 4 \times 2 \times 6 = 49 - 48 = 1

(ii) 3 x^{2} - 2 x + 8 = 0 Here, a = 3, b = - 2, c = 8 Discriminant D is given by : D = b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 8 = 4 - 96 = - 92

(iii) 2 x^{2} - 5 \sqrt{2 x} + 4 = 0 Here, a = 2, b = - 5 \sqrt{2}, c = 4 Discriminant D is given by : D = b^{2} - 4 a c = {(- 5 \sqrt{2})}^{2} - 4 \times 2 \times 4 = (25 \times 2) - 32 = 50 - 32 = 18

(iv) \sqrt{3} x^{2} + 2 \sqrt{2} x - 2 \sqrt{3} = 0 Here, a = \sqrt{3}, b = 2 \sqrt{2}, c = - 2 \sqrt{3} Discriminant D is given by : D = b^{2} - 4 a c = {(2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = (4 \times 2) + (8 \times 3) = 8 + 24 = 32

(v)

(x - 1) (2 x - 1) = 0

\Rightarrow 2 x^{2} - 3 x + 1 = 0

Comparing it with

a x^{2} + b x + c = 0

, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D =

b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 2 \times 1 = 9 - 8 = 1

(vi) 1 - x = 2 x^{2} \Rightarrow 2 x^{2} + x - 1 = 0 Here, a = 2, b = 1, c = - 1 Discriminant D is given by : D = b^{2} - 4 a c {= 1}^{2} - 4 \times 2 (- 1) = 1 + 8 = 9

Page No 191:

Question 2:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 4 x - 1 = 0$

Answer:

$Given : x^{2} - 4 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0, we get : a = 1, b = - 4 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 4)}^{2} - 4 \times 1 \times (- 1) = 16 + 4 = 20 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 4) + \sqrt{20}}{2 \times 1} = \frac{4 + 2 \sqrt{5}}{2} = \frac{2 (2 + \sqrt{5})}{2} = (2 + \sqrt{5}) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 4) - \sqrt{20}}{2} = \frac{4 - 2 \sqrt{5}}{2} = \frac{2 (2 - \sqrt{5})}{2} = (2 - \sqrt{5}) Thus, the roots of the equation are (2 + \sqrt{5}) and (2 - \sqrt{5}) .$

Page No 191:

Question 3:

$x^{2} - 6 x + 4 = 0$

Answer:

$Given : x^{2} - 6 x + 4 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 1, b = - 6 and c = 4 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 1 \times 4 = 36 - 16 = 20 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + \sqrt{20}}{2 \times 1} = \frac{6 + 2 \sqrt{5}}{2} = \frac{2 (3 + \sqrt{5})}{2} = (3 + \sqrt{5}) \begin{array}{l} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - \sqrt{20}}{2 \times 1} = \frac{6 - 2 \sqrt{5}}{2} = \frac{2 (3 - \sqrt{5})}{2} = (3 - \sqrt{5}) \\ Thus, the roots of the equation are (3 + 2 \sqrt{5}) and (3 - 2 \sqrt{5}) . \end{array}$

Page No 191:

Question 4:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 x^{2} + x - 4 = 0$

Answer:

The given equation is $2 x^{2} + x - 4 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = $b^{2} - 4 a c = {(1)}^{2} - 4 \times 2 \times (- 4) = 1 + 32 = 33 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{33}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 1 + \sqrt{33}}{2 \times 2} = \frac{- 1 + \sqrt{33}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 1 - \sqrt{33}}{2 \times 2} = \frac{- 1 - \sqrt{33}}{4}$

Hence, $\frac{- 1 + \sqrt{33}}{4}$ and $\frac{- 1 - \sqrt{33}}{4}$ are the roots of the given equation.

Page No 191:

Question 5:

$25 x^{2} + 30 x + 7 = 0$

Answer:

$Given: 25 x^{2} + 30 x + 7 = 0 On comparing it with a x^{2} + b x + c = 0, we get: a = 25, b = 30 and c = 7 Discriminant D is given by : D = (b^{2} - 4 a c) {= 30}^{2} - 4 \times 25 \times 7 = 900 - 700 = 200 = 200 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 30 + \sqrt{200}}{2 \times 25} = \frac{- 30 + 10 \sqrt{2}}{50} = \frac{10 (- 3 + \sqrt{2})}{50} = \frac{(- 3 + \sqrt{2})}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 30 - \sqrt{200}}{2 \times 25} = \frac{- 30 - 10 \sqrt{2}}{50} = \frac{10 (- 3 - \sqrt{2})}{50} = \frac{(- 3 - \sqrt{2})}{5} Thus, the roots of the equation are \frac{(- 3 + \sqrt{2})}{5} and \frac{(- 3 - \sqrt{2})}{5} .$

Page No 191:

Question 6:

$16 x^{2} = 24 x + 1$

Answer:

$Given : 16 x^{2} = 24 x + 1 ⇒ 16 x^{2} - 24 x - 1 = 0 On c omparing it with a x^{2} + b x + c = 0 a = 16, b = - 24 and c = - 1 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 24)}^{2} - 4 \times 16 \times (- 1) = 576 + (64) = 640 > 0 Hence, the roots of the equation are real, Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 24) + \sqrt{640}}{2 \times 16} = \frac{24 + 8 \sqrt{10}}{32} = \frac{8 (3 + \sqrt{10})}{32} = \frac{(3 + \sqrt{10})}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 24) - \sqrt{640}}{2 \times 16} = \frac{24 - 8 \sqrt{10}}{32} = \frac{8 (3 - \sqrt{10})}{32} = \frac{(3 - \sqrt{10})}{4} Thus, the roots of the equation are \frac{(3 + \sqrt{10})}{4} and \frac{(3 - \sqrt{10})}{4} .$

Page No 191:

Question 7:

$15 x^{2} - 28 = x$

Answer:

$Given: 15 x^{2} - 28 = x ⇒ 15 x^{2} - x - 28 = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = 15, b = - 1 and c = - 28 Discriminant D is given by : D = (b^{2} - 4 a c) = {(- 1)}^{2} - 4 \times 15 \times (- 28) = 1 - (- 1680) = 1 + 1680 = 1681 = 1681 > 0 Hence, the roots of the equation are real. Roots α and β are given by : α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 1) + \sqrt{1681}}{2 \times 15} = \frac{1 + 41}{30} = \frac{42}{30} = \frac{7}{5} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 1) - \sqrt{1681}}{2 \times 15} = \frac{1 - 41}{30} = \frac{- 40}{30} = \frac{- 4}{3} Thus, the roots of the equation are \frac{7}{5} and \frac{- 4}{3} .$

Page No 191:

Question 8:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 x^{2} - 2 \sqrt{2} x + 1 = 0$

Answer:

The given equation is $2 x^{2} - 2 \sqrt{2} x + 1 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $- 2 \sqrt{2}$ and c = 1

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times 2 \times 1 = 8 - 8 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - \sqrt{0}}{2 \times 2} = \frac{2 \sqrt{2}}{4} = \frac{\sqrt{2}}{2}$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

Page No 191:

Question 9:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ [CBSE 2013]

Answer:

The given equation is $\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{2}$ , b = 7 and c = $5 \sqrt{2}$

∴ Discriminant, D = $b^{2} - 4 a c = {(7)}^{2} - 4 \times \sqrt{2} \times 5 \sqrt{2} = 49 - 40 = 9 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9} = 3$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 7 + 3}{2 \times \sqrt{2}} = \frac{- 4}{2 \sqrt{2}} = - \sqrt{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 7 - 3}{2 \times \sqrt{2}} = \frac{- 10}{2 \sqrt{2}} = - \frac{5 \sqrt{2}}{2}$

Hence, $- \sqrt{2}$ and $- \frac{5 \sqrt{2}}{2}$ are the roots of the given equation.

Page No 191:

Question 10:

$\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$

Answer:

$Given : \sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 On c omparing it with a x^{2} + b x + c = 0, we get: a = \sqrt{3}, b = 10 and c = - 8 \sqrt{3} Discriminant D is given by : D = (b^{2} - 4 a c) = {(10)}^{2} - 4 \times \sqrt{3} \times (- 8 \sqrt{3}) = 100 + 96 = 196 > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 10 + \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 10 - \sqrt{196}}{2 \sqrt{3}} = \frac{- 10 - 14}{2 \sqrt{3}} = \frac{- 24}{2 \sqrt{3}} = \frac{- 12}{\sqrt{3}} = \frac{- 12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{- 12 \sqrt{3}}{3} = - 4 \sqrt{3} Thus, the roots of the equation are \frac{2 \sqrt{3}}{3} and - 4 \sqrt{3} .$

Page No 191:

Question 11:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ [CBSE 2015]

Answer:

The given equation is $\sqrt{3} x^{2} - 2 \sqrt{2} x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $\sqrt{3}$ , b = $- 2 \sqrt{2}$ and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{2})}^{2} - 4 \times \sqrt{3} \times (- 2 \sqrt{3}) = 8 + 24 = 32 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{32} = 4 \sqrt{2}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) + 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{6 \sqrt{2}}{2 \sqrt{3}} = \sqrt{6} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{2}) - 4 \sqrt{2}}{2 \times \sqrt{3}} = \frac{- 2 \sqrt{2}}{2 \sqrt{3}} = - \frac{\sqrt{6}}{3}$

Hence, $\sqrt{6}$ and $- \frac{\sqrt{6}}{3}$ are the roots of the given equation.

Page No 191:

Question 12:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 x^{2} + 6 \sqrt{3} x - 60 = 0$ [CBSE 2011, 15]

Answer:

The given equation is $2 x^{2} + 6 \sqrt{3} x - 60 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $6 \sqrt{3}$ and c = $- 60$

∴ Discriminant, D = $b^{2} - 4 a c = {(6 \sqrt{3})}^{2} - 4 \times 2 \times (- 60) = 108 + 480 = 588 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{588} = 14 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} + 14 \sqrt{3}}{2 \times 2} = \frac{8 \sqrt{3}}{4} = 2 \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 6 \sqrt{3} - 14 \sqrt{3}}{2 \times 2} = \frac{- 20 \sqrt{3}}{4} = - 5 \sqrt{3}$

Hence, $2 \sqrt{3}$ and $- 5 \sqrt{3}$ are the roots of the given equation.

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Question 13:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ [CBSE 2013]

Answer:

The given equation is $4 \sqrt{3} x^{2} + 5 x - 2 \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $4 \sqrt{3}$ , b = 5 and c = $- 2 \sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = 5^{2} - 4 \times 4 \sqrt{3} \times (- 2 \sqrt{3}) = 25 + 96 = 121 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{121} = 11$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 + 11}{2 \times 4 \sqrt{3}} = \frac{6}{8 \sqrt{3}} = \frac{\sqrt{3}}{4} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 - 11}{2 \times 4 \sqrt{3}} = \frac{- 16}{8 \sqrt{3}} = - \frac{2 \sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{4}$ and $- \frac{2 \sqrt{3}}{3}$ are the roots of the given equation.

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Question 14:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$3 x^{2} - 2 \sqrt{6} x + 2 = 0$ [CBSE 2012]

Answer:

The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = $- 2 \sqrt{6}$ and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

So, the given equation has real roots.

Now, $\sqrt{D} = 0$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) + 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 \sqrt{6}) - 0}{2 \times 3} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

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Question 15:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 \sqrt{3} x^{2} - 5 x + \sqrt{3} = 0$ [CBSE 2011]

Answer:

The given equation is $2 \sqrt{3} x^{2} - 5 x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = $2 \sqrt{3}$ , b = $- 5$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {(- 5)}^{2} - 4 \times 2 \sqrt{3} \times \sqrt{3} = 25 - 24 = 1 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{1} = 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 5) + 1}{2 \times 2 \sqrt{3}} = \frac{6}{4 \sqrt{3}} = \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 5) - 1}{2 \times 2 \sqrt{3}} = \frac{4}{4 \sqrt{3}} = \frac{\sqrt{3}}{3}$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

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Question 16:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} + x + 2 = 0$

Answer:

The given equation is $x^{2} + x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = 1^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

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Question 17:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 x^{2} + a x - a^{2} = 0$ [CBSE 2015]

Answer:

The given equation is $2 x^{2} + a x - a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 2, B = a and C = $- a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = a^{2} - 4 \times 2 \times - a^{2} = a^{2} + 8 a^{2} = 9 a^{2} \geq 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{9 a^{2}} = 3 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- a + 3 a}{2 \times 2} = \frac{2 a}{4} = \frac{a}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- a - 3 a}{2 \times 2} = \frac{- 4 a}{4} = - a$

Hence, $\frac{a}{2}$ and $- a$ are the roots of the given equation.

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Question 18:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ [CBSE 2015]

Answer:

The given equation is $x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 1, b = $- (\sqrt{3} + 1)$ and c = $\sqrt{3}$

∴ Discriminant, D = $b^{2} - 4 a c = {[- (\sqrt{3} + 1)]}^{2} - 4 \times 1 \times \sqrt{3} = 3 + 1 + 2 \sqrt{3} - 4 \sqrt{3} = 3 - 2 \sqrt{3} + 1 = {(\sqrt{3} - 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(\sqrt{3} - 1)}^{2}} = \sqrt{3} - 1$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] + (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2 \sqrt{3}}{2} = \sqrt{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- [- (\sqrt{3} + 1)] - (\sqrt{3} - 1)}{2 \times 1} = \frac{\sqrt{3} + 1 - \sqrt{3} + 1}{2} = \frac{2}{2} = 1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

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Question 19:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2 x^{2} + 5 \sqrt{3} x + 6 = 0$

Answer:

The given equation is $2 x^{2} + 5 \sqrt{3} x + 6 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 2, b = $5 \sqrt{3}$ and c = 6

∴ Discriminant, D = $b^{2} - 4 a c = {(5 \sqrt{3})}^{2} - 4 \times 2 \times 6 = 75 - 48 = 27 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{27} = 3 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} + 3 \sqrt{3}}{2 \times 2} = \frac{- 2 \sqrt{3}}{4} = - \frac{\sqrt{3}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 5 \sqrt{3} - 3 \sqrt{3}}{2 \times 2} = \frac{- 8 \sqrt{3}}{4} = - 2 \sqrt{3}$

Hence, $- \frac{\sqrt{3}}{2}$ and $- 2 \sqrt{3}$ are the roots of the given equation.

Page No 191:

Question 20:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$3 x^{2} - 2 x + 2 = 0$

Answer:

The given equation is $3 x^{2} - 2 x + 2 = 0$ .

Comparing it with $a x^{2} + b x + c = 0$ , we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2)}^{2} - 4 \times 3 \times 2 = 4 - 24 = - 20 < 0$

Hence, the given equation has no real roots (or real roots does not exist).

Page No 191:

Question 21:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x + \frac{1}{x} = 3, x \neq 0$

Answer:

The given equation is

$x + \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} + 1}{x} = 3 \Rightarrow x^{2} + 1 = 3 x \Rightarrow x^{2} - 3 x + 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times 1 = 9 - 4 = 5 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{5}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{5}}{2 \times 1} = \frac{3 + \sqrt{5}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{5}}{2 \times 1} = \frac{3 - \sqrt{5}}{2}$

Hence, $\frac{3 + \sqrt{5}}{2}$ and $\frac{3 - \sqrt{5}}{2}$ are the roots of the given equation.

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Question 22:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2$ [CBSE 2010]

Answer:

The given equation is

$\frac{1}{x} - \frac{1}{x - 2} = 3, x \neq 0, 2 \Rightarrow \frac{x - 2 - x}{x (x - 2)} = 3 \Rightarrow \frac{- 2}{x^{2} - 2 x} = 3 \Rightarrow - 2 = 3 x^{2} - 6 x$

$\Rightarrow 3 x^{2} - 6 x + 2 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = −6 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 6)}^{2} - 4 \times 3 \times 2 = 36 - 24 = 12 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{12} = 2 \sqrt{3}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 6) + 2 \sqrt{3}}{2 \times 3} = \frac{6 + 2 \sqrt{3}}{6} = \frac{3 + \sqrt{3}}{3} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 6) - 2 \sqrt{3}}{2 \times 3} = \frac{6 - 2 \sqrt{3}}{6} = \frac{3 - \sqrt{3}}{3}$

Hence, $\frac{3 + \sqrt{3}}{3}$ and $\frac{3 - \sqrt{3}}{3}$ are the roots of the given equation.

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Question 23:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x - \frac{1}{x} = 3, x \neq 0$ [CBSE 2010]

Answer:

The given equation is

$x - \frac{1}{x} = 3, x \neq 0 \Rightarrow \frac{x^{2} - 1}{x} = 3 \Rightarrow x^{2} - 1 = 3 x \Rightarrow x^{2} - 3 x - 1 = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −3 and c = −1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 3)}^{2} - 4 \times 1 \times (- 1) = 9 + 4 = 13 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{13}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 3) + \sqrt{13}}{2 \times 1} = \frac{3 + \sqrt{13}}{2} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 3) - \sqrt{13}}{2 \times 1} = \frac{3 - \sqrt{13}}{2}$

Hence, $\frac{3 + \sqrt{13}}{2}$ and $\frac{3 - \sqrt{13}}{2}$ are the roots of the given equation.

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Question 24:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x$

Answer:

The given equation is

$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x \Rightarrow \frac{m^{2} x^{2} + n^{2}}{m n} = 1 - 2 x \Rightarrow m^{2} x^{2} + n^{2} = m n - 2 m n x \Rightarrow m^{2} x^{2} + 2 m n x + n^{2} - m n = 0$

This equation is of the form $a x^{2} + b x + c = 0$ , where a = $m^{2}$ , b = 2mn and c = $n^{2} - m n$ .

∴ Discriminant, D = $b^{2} - 4 a c = {(2 m n)}^{2} - 4 \times m^{2} \times (n^{2} - m n) = 4 m^{2} n^{2} - 4 m^{2} n^{2} + 4 m^{3} n = 4 m^{3} n > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 m^{3} n} = 2 m \sqrt{m n}$

$∴ α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 2 m n + 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{2 m (- n + \sqrt{m n})}{2 m^{2}} = \frac{- n + \sqrt{m n}}{m} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 2 m n - 2 m \sqrt{m n}}{2 \times m^{2}} = \frac{- 2 m (n + \sqrt{m n})}{2 m^{2}} = \frac{- n - \sqrt{m n}}{m}$

Hence, $\frac{- n + \sqrt{m n}}{m}$ and $\frac{- n - \sqrt{m n}}{m}$ are the roots of the given equation.

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Question 25:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$36 x^{2} - 12 a x + (a^{2} - b^{2}) = 0$

Answer:

The given equation is $36 x^{2} - 12 a x + (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 36, B = $- 12 a$ and C = $a^{2} - b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 12 a)}^{2} - 4 \times 36 \times (a^{2} - b^{2}) = 144 a^{2} - 144 a^{2} + 144 b^{2} = 144 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{144 b^{2}} = 12 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 12 a) + 12 b}{2 \times 36} = \frac{12 (a + b)}{72} = \frac{a + b}{6} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 12 a) - 12 b}{2 \times 36} = \frac{12 (a - b)}{72} = \frac{a - b}{6}$

Hence, $\frac{a + b}{6}$ and $\frac{a - b}{6}$ are the roots of the given equation.

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Question 26:

$x^{2} - 2 a x + (a^{2} - b^{2}) = 0$

Answer:

$Given: x^{2} - 2 a x + (a^{2} - b^{2}) = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 1, B = - 2 a and C = (a^{2} - b^{2}) Discriminant D is given by : D = B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - b^{2}) = 4 a^{2} - 4 a^{2} + 4 b^{2} = 4 b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- (- 2 a) + \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a + 2 b}{2} = \frac{2 (a + b)}{2} = (a + b) β = \frac{- b - \sqrt{D}}{2 a} = \frac{- (- 2 a) - \sqrt{4 b^{2}}}{2 \times 1} = \frac{2 a - 2 b}{2} = \frac{2 (a - b)}{2} = (a - b) Hence, the roots of the equation are (a + b) and (a - b) .$

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Question 27:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ [CBSE 2015]

Answer:

The given equation is $x^{2} - 2 a x - (4 b^{2} - a^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- 2 a$ and C = $- (4 b^{2} - a^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 2 a)}^{2} - 4 \times 1 \times [- (4 b^{2} - a^{2})] = 4 a^{2} + 16 b^{2} - 4 a^{2} = 16 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{2}} = 4 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 2 a) + 4 b}{2 \times 1} = \frac{2 (a + 2 b)}{2} = a + 2 b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 2 a) - 4 b}{2 \times 1} = \frac{2 (a - 2 b)}{2} = a - 2 b$

Hence, $a + 2 b$ and $a - 2 b$ are the roots of the given equation.

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Question 28:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ [CBSE 2015]

Answer:

The given equation is $x^{2} + 6 x - (a^{2} + 2 a - 8) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 6 and C = $- (a^{2} + 2 a - 8)$

∴ Discriminant, D = $B^{2} - 4 A C = 6^{2} - 4 \times 1 \times [- (a^{2} + 2 a - 8)] = 36 + 4 a^{2} + 8 a - 32 = 4 a^{2} + 8 a + 4 = 4 (a^{2} + 2 a + 1) = 4 {(a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 {(a + 1)}^{2}} = 2 (a + 1)$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 6 + 2 (a + 1)}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 6 - 2 (a + 1)}{2 \times 1} = \frac{- 2 a - 8}{2} = - a - 4 = - (a + 4)$

Hence, $(a - 2)$ and $- (a + 4)$ are the roots of the given equation.

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Question 29:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} + 5 x - (a^{2} + a - 6) = 0$ [CBSE 2015]

Answer:

The given equation is $x^{2} + 5 x - (a^{2} + a - 6) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = 5 and C = $- (a^{2} + a - 6)$

∴ Discriminant, D = $B^{2} - 4 A C = 5^{2} - 4 \times 1 \times [- (a^{2} + a - 6)] = 25 + 4 a^{2} + 4 a - 24 = 4 a^{2} + 4 a + 1 = {(2 a + 1)}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(2 a + 1)}^{2}} = 2 a + 1$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 5 + 2 a + 1}{2 \times 1} = \frac{2 a - 4}{2} = a - 2 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 5 - (2 a + 1)}{2 \times 1} = \frac{- 2 a - 6}{2} = - a - 3 = - (a + 3)$

Hence, $(a - 2)$ and $- (a + 3)$ are the roots of the given equation.

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Question 30:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ [CBSE 2012]

Answer:

The given equation is $x^{2} - 4 a x - b^{2} + 4 a^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = −4a and C = $- b^{2} + 4 a^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a)}^{2} - 4 \times 1 \times (- b^{2} + 4 a^{2}) = 16 a^{2} + 4 b^{2} - 16 a^{2} = 4 b^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{4 b^{2}} = 2 b$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a) + 2 b}{2 \times 1} = \frac{4 a + 2 b}{2} = 2 a + b β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a) - 2 b}{2 \times 1} = \frac{4 a - 2 b}{2} = 2 a - b$

Hence, $(2 a + b)$ and $(2 a - b)$ are the roots of the given equation.

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Question 31:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ [CBSE 2015]

Answer:

The given equation is $4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = −4a² and C = $a^{4} - b^{4}$

∴ Discriminant, D = $B^{2} - 4 A C = {(- 4 a^{2})}^{2} - 4 \times 4 \times (a^{4} - b^{4}) = 16 a^{4} - 16 a^{4} + 16 b^{4} = 16 b^{4} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 b^{4}} = 4 b^{2}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) + 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} + b^{2})}{8} = \frac{a^{2} + b^{2}}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- (- 4 a^{2}) - 4 b^{2}}{2 \times 4} = \frac{4 (a^{2} - b^{2})}{8} = \frac{a^{2} - b^{2}}{2}$

Hence, $\frac{1}{2} (a^{2} + b^{2})$ and $\frac{1}{2} (a^{2} - b^{2})$ are the roots of the given equation.

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Question 32:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ [CBSE 2015]

Answer:

The given equation is $4 x^{2} + 4 b x - (a^{2} - b^{2}) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 4, B = 4b and C = $- (a^{2} - b^{2})$

∴ Discriminant, D = $B^{2} - 4 A C = {(4 b)}^{2} - 4 \times 4 \times [- (a^{2} - b^{2})] = 16 b^{2} + 16 a^{2} - 16 b^{2} = 16 a^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{16 a^{2}} = 4 a$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- 4 b + 4 a}{2 \times 4} = \frac{4 (a - b)}{8} = \frac{a - b}{2} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- 4 b - 4 a}{2 \times 4} = \frac{- 4 (a + b)}{8} = - \frac{a + b}{2}$

Hence, $\frac{1}{2} (a - b)$ and $- \frac{1}{2} (a + b)$ are the roots of the given equation.

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Question 33:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ [CBSE 2015]

Answer:

The given equation is $x^{2} - (2 b - 1) x + (b^{2} - b - 20) = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = 1, B = $- (2 b - 1)$ and C = $b^{2} - b - 20$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (2 b - 1)]}^{2} - 4 \times 1 \times (b^{2} - b - 20) = 4 b^{2} - 4 b + 1 - 4 b^{2} + 4 b + 80 = 81 > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{81} = 9$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] + 9}{2 \times 1} = \frac{2 b + 8}{2} = b + 4 β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (2 b - 1)] - 9}{2 \times 1} = \frac{2 b - 10}{2} = b - 5$

Hence, $(b + 4)$ and $(b - 5)$ are the roots of the given equation.

Page No 192:

Question 34:

$3 a^{2} x^{2} + 8 a b x + 4 b^{2} = 0, a \neq 0$

Answer:

$Given: {3a}^{2} x^{2} + 8 a b x + 4 b^{2} = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 3 a^{2}, B = 8 a b and C = 4 b^{2} Discriminant D is given by : D = (B^{2} - 4 A C) = {(8 a b)}^{2} - 4 \times 3 a^{2} \times 4 b^{2} {= 16a}^{2} b^{2} > 0 Hence, the roots of the equation are real. Roots α and β are given by: α = \frac{- b + \sqrt{D}}{2 a} = \frac{- 8 a b + \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b + 4 a b}{6 a^{2}} = \frac{- 4 a b}{6 a^{2}} = \frac{- 2 b}{3 a} β = \frac{- b - \sqrt{D}}{2 a} = \frac{- 8 a b - \sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}} = \frac{- 8 a b - 4 a b}{6 a^{2}} = \frac{- 12 a b}{6 a^{2}} = \frac{- 2 b}{a} Thus, the roots of the equation are \frac{- 2 b}{3 a} and \frac{- 2 b}{a} .$

Page No 192:

Question 35:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$ , $a \neq 0$ and $b \neq 0$ [CBSE 2006]

Answer:

The given equation is $a^{2} b^{2} x^{2} - (4 b^{4} - 3 a^{4}) x - 12 a^{2} b^{2} = 0$ .

Comparing it with $A x^{2} + B x + C = 0$ , we get

A = $a^{2} b^{2}$ , B = $- (4 b^{4} - 3 a^{4})$ and C = $- 12 a^{2} b^{2}$

∴ Discriminant, D = $B^{2} - 4 A C = {[- (4 b^{4} - 3 a^{4})]}^{2} - 4 \times a^{2} b^{2} \times (- 12 a^{2} b^{2}) = 16 b^{8} - 24 a^{4} b^{4} + 9 a^{8} + 48 a^{4} b^{4} = 16 b^{8} + 24 a^{4} b^{4} + 9 a^{8} = {(4 b^{4} + 3 a^{4})}^{2} > 0$

So, the given equation has real roots.

Now, $\sqrt{D} = \sqrt{{(4 b^{4} + 3 a^{4})}^{2}} = 4 b^{4} + 3 a^{4}$

$∴ α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (4 b^{4} - 3 a^{4})] + (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{8 b^{4}}{2 a^{2} b^{2}} = \frac{4 b^{2}}{a^{2}} β = \frac{- B - \sqrt{D}}{2 A} = = \frac{- [- (4 b^{4} - 3 a^{4})] - (4 b^{4} + 3 a^{4})}{2 \times a^{2} b^{2}} = \frac{- 6 a^{4}}{2 a^{2} b^{2}} = - \frac{3 a^{2}}{b^{2}}$

Hence, $\frac{4 b^{2}}{a^{2}}$ and $- \frac{3 a^{2}}{b^{2}}$ are the roots of the given equation.

Page No 192:

Question 36:

$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$ , where a $\neq$ 0 and b $\neq$ 0

Answer:

$Given: 12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 On c omparing it with A x^{2} + B x + C = 0, we get: A = 12 a b, B = - (9 a^{2} - 8 b^{2}) and C = - 6 a b Discriminant D is given by : D = B^{2} - 4 A C = {[- (9 a^{2} - 8 b^{2})]}^{2} - 4 \times 12 a b \times (- 6 a b) = 81 a^{4} - 144 a^{2} b^{2} + 64 b^{4} + 288 a^{2} b^{2} = 81 a^{^{4}} + 144 a^{2} b^{2} + 64 b^{4} = {(9 a^{2} + 8 b^{2})}^{2} > 0 Hence, the roots of the equation are equal. Roots α and β are given by: α = \frac{- B + \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] + \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} + 9 a^{2} + 8 b^{2}}{24 a b} = \frac{18 a^{2}}{24 a b} = \frac{3 a}{4 b} β = \frac{- B - \sqrt{D}}{2 A} = \frac{- [- (9 a^{2} - 8 b^{2})] - \sqrt{{(9 a^{2} + 8 b^{2})}^{2}}}{2 \times 12 a b} = \frac{9 a^{2} - 8 b^{2} - 9 a^{2} - 8 b^{2}}{24 a b} = \frac{- 16 b^{2}}{24 a b} = \frac{- 2 b}{3 a} Thus, the roots of the equation are \frac{3 a}{4 b} and \frac{- 2 b}{3 a} .$

Page No 199:

Question 1:

Find the nature of the roots of the following quadratic equations:
(i) $2 x^{2} - 8 x + 5 = 0$
(ii) $3 x^{2} - 2 \sqrt{6} x + 2 = 0$
(iii) $5 x^{2} - 4 x + 1 = 0$
(iv) $5 x (x - 2) + 6 = 0$
(v) $12 x^{2} - 4 \sqrt{15} x + 5 = 0$
(vi) $x^{2} - x + 2 = 0$

Answer:

(i) The given equation is $2 x^{2} - 8 x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2, b = −8 and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 8)}^{2} - 4 \times 2 \times 5 = 64 - 40 = 24 > 0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3 x^{2} - 2 \sqrt{6} x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3, b = $- 2 \sqrt{6}$ and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 2 \sqrt{6})}^{2} - 4 \times 3 \times 2 = 24 - 24 = 0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5 x^{2} - 4 x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −4 and c = 1.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4)}^{2} - 4 \times 5 \times 1 = 16 - 20 = - 4 < 0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5 x (x - 2) + 6 = 0 \Rightarrow 5 x^{2} - 10 x + 6 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = 5, b = −10 and c = 6.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 10)}^{2} - 4 \times 5 \times 6 = 100 - 120 = - 20 < 0$

Hence, the given equation has no real roots.

(v) The given equation is $12 x^{2} - 4 \sqrt{15} x + 5 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 12, b = $- 4 \sqrt{15}$ and c = 5.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 4 \sqrt{15})}^{2} - 4 \times 12 \times 5 = 240 - 240 = 0$

Hence, the given equation has real and equal roots.

(vi) The given equation is $x^{2} - x + 2 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 1, b = −1 and c = 2.

∴ Discriminant, D = $b^{2} - 4 a c = {(- 1)}^{2} - 4 \times 1 \times 2 = 1 - 8 = - 7 < 0$

Hence, the given equation has no real roots.

Page No 199:

Question 2:

If a and b are distinct real numbers, show that the quadratic equation $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ has no real roots.

Answer:

The given equation is $2 (a^{2} + b^{2}) x^{2} + 2 (a + b) x + 1 = 0$ .
$∴ D = {[2 (a + b)]}^{2} - 4 \times 2 (a^{2} + b^{2}) \times 1 = 4 (a^{2} + 2 a b + b^{2}) - 8 (a^{2} + b^{2}) = 4 a^{2} + 8 a b + 4 b^{2} - 8 a^{2} - 8 b^{2} = - 4 a^{2} + 8 a b - 4 b^{2} = - 4 (a^{2} - 2 a b + b^{2}) = - 4 {(a - b)}^{2} < 0$
Hence, the given equation has no real roots.

Page No 199:

Question 3:

Show that the roots of the equation $x^{2} + p x - q^{2} = 0$ are real for all real value of p and q.

Answer:

$Given: x^{2} + p x - q^{2} = 0 Here, a = 1, b = p and c = - q^{2} Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 1 \times (- q^{2}) = (p^{2} + 4 q^{2}) > 0 D > 0 for all real values of p and q . Thus, the roots of the equation are real .$

Page No 199:

Question 4:

For what values of k are the roots of the quadratic equation $3 x^{2} + 2 k x + 27 = 0$ real and equal?

Answer:

$Given: 3 x^{2} + 2 k x + 27 = 0 Here, a = 3, b = 2 k and c = 27 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 3 \times 27 = 0 \Rightarrow 4 k^{2} - 324 = 0 \Rightarrow 4 k^{2} = 324 \Rightarrow k^{2} = 81 \Rightarrow k = \pm 9 ∴ k = 9 or k = - 9$

Page No 199:

Question 5:

For what value of k are the roots of the quadratic equation $k x (x - 2 \sqrt{5}) + 10 = 0$ real and equal? [CBSE 2013]

Answer:

The given equation is

$k x (x - 2 \sqrt{5}) + 10 = 0 \Rightarrow k x^{2} - 2 \sqrt{5} k x + 10 = 0$

This is of the form $a x^{2} + b x + c = 0$ , where a = k, b = $- 2 \sqrt{5} k$ and c = 10.

$∴ D = b^{2} - 4 a c = {(- 2 \sqrt{5} k)}^{2} - 4 \times k \times 10 = 20 k^{2} - 40 k$

The given equation will have real and equal roots if D = 0.

$∴ 20 k^{2} - 40 k = 0 \Rightarrow 20 k (k - 2) = 0 \Rightarrow k = 0 or k - 2 = 0 \Rightarrow k = 0 or k = 2$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

Page No 199:

Question 6:

For what values of p are the roots of the equation $4 x^{2} + p x + 3 = 0$ real and equal? [CBSE 2014]

Answer:

The given equation is $4 x^{2} + p x + 3 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 4, b = p and c = 3.

$∴ D = b^{2} - 4 a c = p^{2} - 4 \times 4 \times 3 = p^{2} - 48$

The given equation will have real and equal roots if D = 0.

$∴ p^{2} - 48 = 0 \Rightarrow p^{2} = 48 \Rightarrow p = \pm \sqrt{48} = \pm 4 \sqrt{3}$

Hence, $4 \sqrt{3}$ and $- 4 \sqrt{3}$ are the required values of p.

Page No 200:

Question 7:

Find the non-zero value of k for which the roots of the quadratic equation $9 x^{2} - 3 k x + k = 0$ are real and equal. [CBSE 2014]

Answer:

The given equation is $9 x^{2} - 3 k x + k = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 9, b = −3k and c = k.

$∴ D = b^{2} - 4 a c = {(- 3 k)}^{2} - 4 \times 9 \times k = 9 k^{2} - 36 k$

The given equation will have real and equal roots if D = 0.

$∴ 9 k^{2} - 36 k = 0 \Rightarrow 9 k (k - 4) = 0 \Rightarrow k = 0 or k - 4 = 0 \Rightarrow k = 0 or k = 4$

But, k ≠ 0 (Given)

Hence, the required value of k is 4.

Page No 200:

Question 8:

(i) Find the values of k for which the quadratic equation $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ has real and equal roots. [CBSE 2014]
(ii) Find the value of k for which the equation $x^{2} + k (2 x + k - 1) + 2 = 0$ has real and equal roots. [CBSE 2017]

Answer:

(i)
The given equation is $(3 k + 1) x^{2} + 2 (k + 1) x + 1 = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 3k +1, b = 2(k + 1) and c = 1.

$∴ D = b^{2} - 4 a c = {[2 (k + 1)]}^{2} - 4 \times (3 k + 1) \times 1 = 4 (k^{2} + 2 k + 1) - 4 (3 k + 1) = 4 k^{2} + 8 k + 4 - 12 k - 4$
$= 4 k^{2} - 4 k$

The given equation will have real and equal roots if D = 0.

$∴ 4 k^{2} - 4 k = 0 \Rightarrow 4 k (k - 1) = 0 \Rightarrow k = 0 or k - 1 = 0 \Rightarrow k = 0 or k = 1$

Hence, 0 and 1 are the required values of k.

(ii)
$x^{2} + k (2 x + k - 1) + 2 = 0 x^{2} + 2 k x + k^{2} - k + 2 = 0 For real and equal roots D = 0 D = b^{2} - 4 a c = 0 D = {(2 k)}^{2} - 4 (1) (k^{2} - k + 2) D = 4 k^{2} - 4 k^{2} + 4 k - 8 = 0 \Rightarrow 4 k - 8 = 0 \Rightarrow k = 2$

Page No 200:

Question 9:

Find the values of p for which the quadratic equation $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ has real and equal roots. [CBSE 2014]

Answer:

The given equation is $(2 p + 1) x^{2} - (7 p + 2) x + (7 p - 3) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = 2p +1, b = −(7p + 2) and c = 7p − 3.

$∴ D = b^{2} - 4 a c = {[- (7 p + 2)]}^{2} - 4 \times (2 p + 1) \times (7 p - 3) = (49 p^{2} + 28 p + 4) - 4 (14 p^{2} + p - 3) = 49 p^{2} + 28 p + 4 - 56 p^{2} - 4 p + 12$
$= - 7 p^{2} + 24 p + 16$

The given equation will have real and equal roots if D = 0.

$∴ - 7 p^{2} + 24 p + 16 = 0 \Rightarrow 7 p^{2} - 24 p - 16 = 0 \Rightarrow 7 p^{2} - 28 p + 4 p - 16 = 0 \Rightarrow 7 p (p - 4) + 4 (p - 4) = 0$
$\Rightarrow (p - 4) (7 p + 4) = 0 \Rightarrow p - 4 = 0 or 7 p + 4 = 0 \Rightarrow p = 4 or p = - \frac{4}{7}$

Hence, 4 and $- \frac{4}{7}$ are the required values of p.

Page No 200:

Question 10:

Find the values of p for which the quadratic equation

(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0, p \neq - 1

has equal roots. Hence, find the roots of the equation. [CBSE 2015]

Answer:

The given equation is $(p + 1) x^{2} - 6 (p + 1) x + 3 (p + 9) = 0$ .

This is of the form $a x^{2} + b x + c = 0$ , where a = p +1, b = −6(p + 1) and c = 3(p + 9).

$∴ D = b^{2} - 4 a c = {[- 6 (p + 1)]}^{2} - 4 \times (p + 1) \times 3 (p + 9) = 12 (p + 1) [3 (p + 1) - (p + 9)] = 12 (p + 1) (2 p - 6)$

The given equation will have real and equal roots if D = 0.

$∴ 12 (p + 1) (2 p - 6) = 0 \Rightarrow p + 1 = 0 or 2 p - 6 = 0 \Rightarrow p = - 1 or p = 3$

But, $p \neq - 1$ (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4 x^{2} - 24 x + 36 = 0$ .

$4 x^{2} - 24 x + 36 = 0 \Rightarrow 4 (x^{2} - 6 x + 9) = 0 \Rightarrow {(x - 3)}^{2} = 0 \Rightarrow x - 3 = 0$
$\Rightarrow x = 3$

Hence, 3 is the repeated root of this equation.

Page No 200:

Question 11:

If −5 is a root of the quadratic equation $2 x^{2} + p x - 15 = 0$ and the quadratic equation $p (x^{2} + x) + k = 0$ has equal roots, find the value of k.
[CBSE 2014]

Answer:

It is given that −5 is a root of the quadratic equation $2 x^{2} + p x - 15 = 0$ .

$∴ 2 {(- 5)}^{2} + p \times (- 5) - 15 = 0 \Rightarrow - 5 p + 35 = 0 \Rightarrow p = 7$

The roots of the equation $p x^{2} + p x + k = 0$ = 0 are equal.

$∴ D = 0 \Rightarrow p^{2} - 4 p k = 0 \Rightarrow {(7)}^{2} - 4 \times 7 \times k = 0 \Rightarrow 49 - 28 k = 0$
$\Rightarrow k = \frac{49}{28} = \frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$ .

Page No 200:

Question 12:

If 3 is a root of the quadratic equation $x^{2} - x + k = 0$ , find the value of p so that the roots of the equation $x^{2} + k (2 x + k + 2) + p = 0$ are equal.
[CBSE 2015]

Answer:

It is given that 3 is a root of the quadratic equation $x^{2} - x + k = 0$ .

$∴ {(3)}^{2} - 3 + k = 0 \Rightarrow k + 6 = 0 \Rightarrow k = - 6$

The roots of the equation $x^{2} + 2 k x + (k^{2} + 2 k + p) = 0$ are equal.

$∴ D = 0 \Rightarrow {(2 k)}^{2} - 4 \times 1 \times (k^{2} + 2 k + p) = 0 \Rightarrow 4 k^{2} - 4 k^{2} - 8 k - 4 p = 0 \Rightarrow - 8 k - 4 p = 0$
$\Rightarrow p = \frac{8 k}{- 4} = - 2 k \Rightarrow p = - 2 \times (- 6) = 12$

Hence, the value of p is 12.

Page No 200:

Question 13:

If −4 is a root of the quadratic equation $x^{2} + 2 x + 4 p = 0$ , find the value of k for which the quadratic equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots. [CBSE 2015]

Answer:

It is given that −4 is a root of the quadratic equation $x^{2} + 2 x + 4 p = 0$ .

$∴ {(- 4)}^{2} + 2 \times (- 4) + 4 p = 0 \Rightarrow 16 - 8 + 4 p = 0 \Rightarrow 4 p + 8 = 0 \Rightarrow p = - 2$

The equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots.

$∴ D = 0 \Rightarrow {[p (1 + 3 k)]}^{2} - 4 \times 1 \times 7 (3 + 2 k) = 0 \Rightarrow {[- 2 (1 + 3 k)]}^{2} - 28 (3 + 2 k) = 0 \Rightarrow 4 (1 + 6 k + 9 k^{2}) - 28 (3 + 2 k) = 0$
$\Rightarrow 4 (1 + 6 k + 9 k^{2} - 21 - 14 k) = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0$
$\Rightarrow (k - 2) (9 k + 10) = 0 \Rightarrow k - 2 = 0 or 9 k + 10 = 0 \Rightarrow k = 2 or k = - \frac{10}{9}$

Hence, the required value of k is 2 or $- \frac{10}{9}$ .

Page No 200:

Question 14:

If the equation $(1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0$ has equal roots, prove that $c^{2} = a^{2} (1 + m^{2}) .$

Answer:

$Given: (1 + m^{2}) x^{2} + 2 m c x + (c^{2} - a^{2}) = 0 Here, a = (1 + m^{2}), b = 2 m c and c = (c^{2} - a^{2}) It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(2 m c)}^{2} - 4 \times (1 + m^{2}) \times (c^{2} - a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 (c^{2} - a^{2} + m^{2} c^{2} - m^{2} a^{2}) = 0 \Rightarrow 4 m^{2} c^{2} - 4 c^{2} + 4 a^{2} - 4 m^{2} c^{2} + 4 m^{2} a^{2} = 0 \Rightarrow - 4 c^{2} + 4 a^{2} + 4 m^{2} a^{2} = 0 \Rightarrow a^{2} + m^{2} a^{2} = c^{2} \Rightarrow a^{2} (1 + m^{2}) = c^{2} \Rightarrow c^{2} = a^{2} (1 + m^{2}) Hence proved .$

Page No 200:

Question 15:

If the roots of the equation $(c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0$ are real and equal, show that either $a = 0 or (a^{3} + b^{3} + c^{3}) = 3 a b c .$

Answer:

$Given: (c^{2} - a b) x^{2} - 2 (a^{2} - b c) x + (b^{2} - a c) = 0 Here, a = (c^{2} - a b), b = - 2 (a^{2} - b c), c = (b^{2} - a c) It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {- 2 (a^{2} - b c)}^{2} - 4 \times (c^{2} - a b) \times (b^{2} - a c) = 0 \Rightarrow 4 (a^{4} - 2 a^{2} b c + b^{2} c^{2}) - 4 (b^{2} c^{2} - a c^{3} - a b^{3} + a^{2} b c) = 0 \Rightarrow a^{4} - 2 a^{2} b c + b^{2} c^{2} - b^{2} c^{2} + a c^{3} + a b^{3} - a^{2} b c = 0 \Rightarrow a^{4} - 3 a^{2} b c + a c^{3} + a b^{3} = 0 \Rightarrow a (a^{3} - 3 a b c + c^{3} + b^{3}) = 0 Now, a = 0 or a^{3} - 3 a b c + c^{3} + b^{3} = 0 a = 0 {or a}^{3} + b^{3} + c^{3} = 3 a b c$

Page No 200:

Question 16:

Determine the values of p for which the quadratic equation $2 x^{2} + p x + 8 = 0$ has real roots.

Answer:

$Given: 2 x^{2} + p x + 8 = 0 Here, a = 2, b = p and c = 8 Discriminant D is given by : D = (b^{2} - 4 a c) = p^{2} - 4 \times 2 \times 8 = (p^{2} - 64) If D \geq 0, t he roots of the equation will be real. \Rightarrow (p^{2} - 64) \geq 0 \Rightarrow (p + 8) (p - 8) \geq 0 \Rightarrow p \geq 8 and p \leq - 8 T hus, the roots of the equation are real for p \geq 8 and p \leq - 8 .$

Page No 200:

Question 17:

Find the value of α for which the equation $(α - 12) x^{2} + 2 (α - 12) x + 2 = 0$ has equal roots.

Answer:

$Given: (α - 12) x^{2} + 2 (α - 12) x + 2 = 0 Here, a = (α - 12), b = 2 (α - 12) and c = 2 It is given that the roots of the equation are equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {2 (α - 12)}^{2} - 4 \times (α - 12) \times 2 = 0 \Rightarrow 4 (α^{2} - 24 α + 144) - 8 (α - 12) = 0 \Rightarrow 4 α^{2} - 96 α + 576 - 8 α + 96 = 0 \Rightarrow 4 α^{2} - 104 α + 672 = 0 \Rightarrow α^{2} - 26 α + 168 = 0 \Rightarrow α^{2} - 14 α - 12 α + 168 = 0 \Rightarrow α (α - 14) - 12 (α - 14) = 0 \Rightarrow (α - 14) (α - 12) = 0 ∴ α = 14 or α = 12 If the value of α is 12, the given equation becomes non-quadratic. Therefore, the valueof α will be 14 for the equation to have equal roots.$

Page No 200:

Question 18:

Find the values k for which of roots of $9 x^{2} + 8 k x + 16 = 0$ are real and equal

Answer:

$Given: 9 x^{2} + 8 k x + 16 = 0 Here, a = 9, b = 8 k and c = 16 It is given that the roots of the equation are real and equal; therefore, we have: D = 0 \Rightarrow (b^{2} - 4 a c) = 0 \Rightarrow {(8 k)}^{2} - 4 \times 9 \times 16 = 0 \Rightarrow 64 k^{2} - 576 = 0 \Rightarrow 64 k^{2} = 576 \Rightarrow k^{2} = 9 \Rightarrow k = \pm 3 ∴ k = 3 or k = - 3$

Page No 200:

Question 19:

Find the values of k for which the given quadratic equation has real and distinct roots:

(i) $k x^{2} + 6 x + 1 = 0$
(ii) $x^{2} - k x + 9 = 0$
(iii) $9 x^{2} + 3 k x + 4 = 0$
(iv) $5 x^{2} - k x + 1 = 0$

Answer:

(i) The given equation is $k x^{2} + 6 x + 1 = 0$ .

$∴ D = 6^{2} - 4 \times k \times 1 = 36 - 4 k$

The given equation has real and distinct roots if D > 0.

$∴ 36 - 4 k > 0 \Rightarrow 4 k < 36 \Rightarrow k < 9$

(ii) The given equation is $x^{2} - k x + 9 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 1 \times 9 = k^{2} - 36$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 36 > 0 \Rightarrow (k - 6) (k + 6) > 0 \Rightarrow k < - 6 or k > 6$

(iii) The given equation is $9 x^{2} + 3 k x + 4 = 0$ .

$∴ D = {(3 k)}^{2} - 4 \times 9 \times 4 = 9 k^{2} - 144$

The given equation has real and distinct roots if D > 0.

$∴ 9 k^{2} - 144 > 0 \Rightarrow 9 (k^{2} - 16) > 0 \Rightarrow (k - 4) (k + 4) > 0 \Rightarrow k < - 4 or k > 4$

(iv) The given equation is $5 x^{2} - k x + 1 = 0$ .

$∴ D = {(- k)}^{2} - 4 \times 5 \times 1 = k^{2} - 20$

The given equation has real and distinct roots if D > 0.

$∴ k^{2} - 20 > 0 \Rightarrow k^{2} - {(2 \sqrt{5})}^{2} > 0 \Rightarrow (k - 2 \sqrt{5}) (k + 2 \sqrt{5}) > 0 \Rightarrow k < - 2 \sqrt{5} or k > 2 \sqrt{5}$

Page No 200:

Question 20:

If a and b are real and a ≠ b then show that the roots of the equation $(a - b) x^{2} + 5 (a + b) x - 2 (a - b) = 0$ are real and unequal.

Answer:

The given equation is $(a - b) x^{2} + 5 (a + b) x - 2 (a - b) = 0$ .

$∴ D = {[5 (a + b)]}^{2} - 4 \times (a - b) \times [- 2 (a - b)] = 25 {(a + b)}^{2} + 8 {(a - b)}^{2}$
Since a and b are real and a ≠ b, so ${(a - b)}^{2} > 0$ and ${(a + b)}^{2} > 0$ .

∴ $8 {(a - b)}^{2} > 0$ .....(1)               (Product of two positive numbers is always positive)

Also, $25 {(a + b)}^{2} > 0$              .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25 {(a + b)}^{2} + 8 {(a - b)}^{2} > 0$                  (Sum of two positive numbers is always positive)

$\Rightarrow D > 0$

Hence, the roots of the given equation are real and unequal.

Page No 201:

Question 21:

(i) If the roots of the equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ are equal, prove that $\frac{a}{b} = \frac{c}{d}$ . [CBSE 2017]
(ii) If ad ≠ bc then prove that the equation $(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ has no real roots. [CBSE 2017]

Answer:

(i)
It is given that the roots of the equation $(a^{2} + b^{2}) x^{2} - 2 (a c + b d) x + (c^{2} + d^{2}) = 0$ are equal.
$∴ D = 0 \Rightarrow {[- 2 (a c + b d)]}^{2} - 4 (a^{2} + b^{2}) (c^{2} + d^{2}) = 0 \Rightarrow 4 (a^{2} c^{2} + b^{2} d^{2} + 2 a b c d) - 4 (a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2}) = 0 \Rightarrow 4 (a^{2} c^{2} + b^{2} d^{2} + 2 a b c d - a^{2} c^{2} - a^{2} d^{2} - b^{2} c^{2} - b^{2} d^{2}) = 0$
$\Rightarrow (- a^{2} d^{2} + 2 a b c d - b^{2} c^{2}) = 0 \Rightarrow - (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}) = 0 \Rightarrow {(a d - b c)}^{2} = 0$
$\Rightarrow a d - b c = 0 \Rightarrow a d = b c \Rightarrow \frac{a}{b} = \frac{c}{d}$
Hence Proved.

(ii)
Consider the equation $(a^{2} + b^{2}) x^{2} + 2 (a c + b d) x + (c^{2} + d^{2}) = 0$
On comparing the above equation with the general equation $a x^{2} + b x + c$ , we get
$a = (a^{2} + b^{2}), b = 2 (a c + b d), c = (c^{2} + d^{2})$
We know that $D = b^{2} - 4 a c$
$\Rightarrow D = {\{2 (a c + b d)\}}^{2} - 4 \times (a^{2} + b^{2}) \times (c^{2} + d^{2}) \Rightarrow D = \{4 {(a c + b d)}^{2}\} - 4 \times (a^{2} + b^{2}) \times (c^{2} + d^{2}) \Rightarrow D = 4 \{a^{2} c^{2} + b^{2} d^{2} + 2 a b c d\} - 4 \{a^{2} c^{2} + a^{2} d^{2} + b^{2} c^{2} + b^{2} d^{2}\} \Rightarrow D = 4 \{a^{2} c^{2} + b^{2} d^{2} + 2 a b c d - a^{2} c^{2} - a^{2} d^{2} - b^{2} c^{2} - b^{2} d^{2}\} \Rightarrow D = 4 \{- a^{2} d^{2} - b^{2} c^{2} + 2 a b c d\} \Rightarrow D = - 4 {(a d - b c)}^{2}$
Since, the value of D is negative, the given equation has no real roots.

Page No 201:

Question 22:

If the roots of the equations $a x^{2} + 2 b x + c = 0$ and $b x^{2} - 2 \sqrt{a c} x + b = 0$ are simultaneously real then prove that $b^{2} = a c$ .

Answer:

It is given that the roots of the equation $a x^{2} + 2 b x + c = 0$ are real.
$∴ D_{1} = {(2 b)}^{2} - 4 \times a \times c \geq 0 \Rightarrow 4 (b^{2} - a c) \geq 0 \Rightarrow b^{2} - a c \geq 0 . . . . . (1)$

Also, the roots of the equation $b x^{2} - 2 \sqrt{a c} x + b = 0$ are real.
$∴ D_{2} = {(- 2 \sqrt{a c})}^{2} - 4 \times b \times b \geq 0 \Rightarrow 4 (a c - b^{2}) \geq 0 \Rightarrow - 4 (b^{2} - a c) \geq 0$
$\Rightarrow b^{2} - a c \leq 0 . . . . . (2)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
$b^{2} - a c = 0 \Rightarrow b^{2} = a c$

Page No 221:

Question 1:

The sum of a natural number and its square is 156. Find the number.

Answer:

Let the required natural number be x.

According to the given condition,

$x + x^{2} = 156 \Rightarrow x^{2} + x - 156 = 0 \Rightarrow x^{2} + 13 x - 12 x - 156 = 0 \Rightarrow x (x + 13) - 12 (x + 13) = 0$
$\Rightarrow (x + 13) (x - 12) = 0 \Rightarrow x + 13 = 0 or x - 12 = 0 \Rightarrow x = - 13 or x = 12$

∴ x = 12 (x cannot be negative)

Hence, the required natural number is 12.

Page No 221:

Question 2:

The sum of a natural number and its positive square root is 132. Find the number.

Answer:

Let the required natural number be x.

According to the given condition,

x + \sqrt{x} = 132

Putting

\sqrt{x} = y

or x = y², we get

y^{2} + y = 132 \Rightarrow y^{2} + y - 132 = 0 \Rightarrow y^{2} + 12 y - 11 y - 132 = 0 \Rightarrow y (y + 12) - 11 (y + 12) = 0

\Rightarrow (y + 12) (y - 11) = 0 \Rightarrow y + 12 = 0 or y - 11 = 0 \Rightarrow y = - 12 or y = 11

∴ y = 11 (y cannot be negative)

Now,

\sqrt{x} = 11 \Rightarrow x = {(11)}^{2} = 121

Hence, the required natural number is 121.

Page No 221:

Question 3:

The sum of two natural numbers is 28 and their product is 192. Find the numbers.

Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

$x (28 - x) = 192 \Rightarrow 28 x - x^{2} = 192 \Rightarrow x^{2} - 28 x + 192 = 0 \Rightarrow x^{2} - 16 x - 12 x + 192 = 0$
$\Rightarrow x (x - 16) - 12 (x - 16) = 0 \Rightarrow (x - 12) (x - 16) = 0 \Rightarrow x - 12 = 0 or x - 16 = 0 \Rightarrow x = 12 or x = 16$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Page No 221:

Question 4:

The sum of the squares of two consecutive positive integers is 365. Find the integers.

Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
$x^{2} + {(x + 1)}^{2} = 365 \Rightarrow x^{2} + x^{2} + 2 x + 1 = 365 \Rightarrow 2 x^{2} + 2 x - 364 = 0 \Rightarrow x^{2} + x - 182 = 0$
$\Rightarrow x^{2} + 14 x - 13 x - 182 = 0 \Rightarrow x (x + 14) - 13 (x + 14) = 0 \Rightarrow (x + 14) (x - 13) = 0 \Rightarrow x + 14 = 0 or x - 13 = 0$
$\Rightarrow x = - 14 or x = 13$
∴ x = 13 (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

Page No 221:

Question 5:

The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.

Answer:

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,
$x^{2} + {(x + 2)}^{2} = 514 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 514 \Rightarrow 2 x^{2} + 4 x - 510 = 0 \Rightarrow x^{2} + 2 x - 255 = 0$
$\Rightarrow x^{2} + 17 x - 15 x - 255 = 0 \Rightarrow x (x + 17) - 15 (x + 17) = 0 \Rightarrow (x + 17) (x - 15) = 0 \Rightarrow x + 17 = 0 or x - 15 = 0$
$\Rightarrow x = - 17 or x = 15$
∴ x = 15 (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.

Page No 221:

Question 6:

The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.

Answer:

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,
$x^{2} + {(x + 2)}^{2} = 452 \Rightarrow x^{2} + x^{2} + 4 x + 4 = 452 \Rightarrow 2 x^{2} + 4 x - 448 = 0 \Rightarrow x^{2} + 2 x - 224 = 0$
$\Rightarrow x^{2} + 16 x - 14 x - 224 = 0 \Rightarrow x (x + 16) - 14 (x + 16) = 0 \Rightarrow (x + 16) (x - 14) = 0 \Rightarrow x + 16 = 0 or x - 14 = 0$
$\Rightarrow x = - 16 or x = 14$
∴ x = 14 (x is a positive even number)

When x = 14,
x + 2 = 14 + 2 = 16

Hence, the required numbers are 14 and 16.

Page No 221:

Question 7:

The product of two consecutive positive integers is 306. Find the integers.

Answer:

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,
$x (x + 1) = 306 \Rightarrow x^{2} + x - 306 = 0 \Rightarrow x^{2} + 18 x - 17 x - 306 = 0 \Rightarrow x (x + 18) - 17 (x + 18) = 0$
$\Rightarrow (x + 18) (x - 17) = 0 \Rightarrow x + 18 = 0 or x - 17 = 0 \Rightarrow x = - 18 or x = 17$
∴ x = 17 (x is a positive integer)

When x = 17,
x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

Page No 221:

Question 8:

Two natural numbers differ by 3 and their product is 504. Find the numbers.

Answer:

$Let the required numbers be x and (x + 3) . According to the question: x (x + 3) = 504 \Rightarrow x^{2} + 3 x = 504 \Rightarrow x^{2} + 3 x - 504 = 0 \Rightarrow x^{2} + (24 - 21) x - 504 = 0 \Rightarrow x^{2} + 24 x - 21 x - 504 = 0 \Rightarrow x (x + 24) - 21 (x + 24) = 0 \Rightarrow (x + 24) (x - 21) = 0 \Rightarrow x + 24 = 0 or x - 21 = 0 \Rightarrow x = - 24 or x = 21 If x = - 24, the numbers are - 24 and {(- 24 + 3) = - 21}. If x = 21, the numbers are 21 and {(21 + 3) = 24}. Hence, the numbers are (- 24, - 21) and (21, 24) .$

Page No 221:

Question 9:

Find two consecutive multiples of 3 whose product is 648.

Answer:

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
$3 x \times 3 (x + 1) = 648 \Rightarrow 9 (x^{2} + x) = 648 \Rightarrow x^{2} + x = 72 \Rightarrow x^{2} + x - 72 = 0$
$\Rightarrow x^{2} + 9 x - 8 x - 72 = 0 \Rightarrow x (x + 9) - 8 (x + 9) = 0 \Rightarrow (x + 9) (x - 8) = 0 \Rightarrow x + 9 = 0 or x - 8 = 0$
$\Rightarrow x = - 9 or x = 8$
∴ x = 8 (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

Page No 221:

Question 10:

Find two consecutive positive odd integers whose product is 483.

Answer:

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,
$x (x + 2) = 483 \Rightarrow x^{2} + 2 x - 483 = 0 \Rightarrow x^{2} + 23 x - 21 x - 483 = 0 \Rightarrow x (x + 23) - 21 (x + 23) = 0$
$\Rightarrow (x + 23) (x - 21) = 0 \Rightarrow x + 23 = 0 or x - 21 = 0 \Rightarrow x = - 23 or x = 21$
∴ x = 21 (x is a positive odd integer)

When x = 21,
x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23.

Page No 221:

Question 11:

Find two consecutive positive even integers whose product is 288.

Answer:

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
$x (x + 2) = 288 \Rightarrow x^{2} + 2 x - 288 = 0 \Rightarrow x^{2} + 18 x - 16 x - 288 = 0 \Rightarrow x (x + 18) - 16 (x + 18) = 0$
$\Rightarrow (x + 18) (x - 16) = 0 \Rightarrow x + 18 = 0 or x - 16 = 0 \Rightarrow x = - 18 or x = 16$
∴ x = 16 (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

Page No 222:

Question 12:

The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}$ . Find the numbers. [CBSE 2012]

Answer:

Let the required natural numbers be x and (9 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{9 - x} = \frac{1}{2} \Rightarrow \frac{9 - x + x}{x (9 - x)} = \frac{1}{2} \Rightarrow \frac{9}{9 x - x^{2}} = \frac{1}{2} \Rightarrow 9 x - x^{2} = 18$
$\Rightarrow x^{2} - 9 x + 18 = 0 \Rightarrow x^{2} - 6 x - 3 x + 18 = 0 \Rightarrow x (x - 6) - 3 (x - 6) = 0 \Rightarrow (x - 3) (x - 6) = 0$
$\Rightarrow x - 3 = 0 or x - 6 = 0 \Rightarrow x = 3 or x = 6$

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.

Page No 222:

Question 13:

The sum of two natural numbers is 15 and the sum of their reciprocals is $\frac{3}{10}$ . Find the numbers. [CBSE 2005]

Answer:

Let the required natural numbers be x and (15 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \Rightarrow \frac{15 - x + x}{x (15 - x)} = \frac{3}{10} \Rightarrow \frac{15}{15 x - x^{2}} = \frac{3}{10} \Rightarrow 15 x - x^{2} = 50$
$\Rightarrow x^{2} - 15 x + 50 = 0 \Rightarrow x^{2} - 10 x - 5 x + 50 = 0 \Rightarrow x (x - 10) - 5 (x - 10) = 0 \Rightarrow (x - 5) (x - 10) = 0$
$\Rightarrow x - 5 = 0 or x - 10 = 0 \Rightarrow x = 5 or x = 10$

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.

Page No 222:

Question 14:

The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$ . Find the numbers. [CBSE 2014]

Answer:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
$∴ \frac{1}{x} > \frac{1}{x + 3}$

According to the given condition,
$\frac{1}{x} - \frac{1}{x + 3} = \frac{3}{28} \Rightarrow \frac{x + 3 - x}{x (x + 3)} = \frac{3}{28} \Rightarrow \frac{3}{x^{2} + 3 x} = \frac{3}{28} \Rightarrow x^{2} + 3 x = 28$
$\Rightarrow x^{2} + 3 x - 28 = 0 \Rightarrow x^{2} + 7 x - 4 x - 28 = 0 \Rightarrow x (x + 7) - 4 (x + 7) = 0 \Rightarrow (x + 7) (x - 4) = 0$
$\Rightarrow x + 7 = 0 or x - 4 = 0 \Rightarrow x = - 7 or x = 4$
∴ x = 4 (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.

Page No 222:

Question 15:

The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}$ . Find the numbers. [CBSE 2014]

Answer:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
$∴ \frac{1}{x} > \frac{1}{x + 5}$

According to the given condition,
$\frac{1}{x} - \frac{1}{x + 5} = \frac{5}{14} \Rightarrow \frac{x + 5 - x}{x (x + 5)} = \frac{5}{14} \Rightarrow \frac{5}{x^{2} + 5 x} = \frac{5}{14} \Rightarrow x^{2} + 5 x = 14$
$\Rightarrow x^{2} + 5 x - 14 = 0 \Rightarrow x^{2} + 7 x - 2 x - 14 = 0 \Rightarrow x (x + 7) - 2 (x + 7) = 0 \Rightarrow (x + 7) (x - 2) = 0$
$\Rightarrow x + 7 = 0 or x - 2 = 0 \Rightarrow x = - 7 or x = 2$
∴ x = 2 (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.

Page No 222:

Question 16:

The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.

Answer:

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,
${(7 x)}^{2} + {[7 (x + 1)]}^{2} = 1225 \Rightarrow 49 x^{2} + 49 (x^{2} + 2 x + 1) = 1225 \Rightarrow 49 x^{2} + 49 x^{2} + 98 x + 49 = 1225 \Rightarrow 98 x^{2} + 98 x - 1176 = 0$
$\Rightarrow x^{2} + x - 12 = 0 \Rightarrow x^{2} + 4 x - 3 x - 12 = 0 \Rightarrow x (x + 4) - 3 (x + 4) = 0 \Rightarrow (x + 4) (x - 3) = 0$
$\Rightarrow x + 4 = 0 or x - 3 = 0 \Rightarrow x = - 4 or x = 3$
∴ x = 3 (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.

Page No 222:

Question 17:

The sum of a natural number and its reciprocal is $\frac{65}{8}$ . Find the number.

Answer:

Let the natural number be x.

According to the given condition,
$x + \frac{1}{x} = \frac{65}{8} \Rightarrow \frac{x^{2} + 1}{x} = \frac{65}{8} \Rightarrow 8 x^{2} + 8 = 65 x \Rightarrow 8 x^{2} - 65 x + 8 = 0$
$\Rightarrow 8 x^{2} - 64 x - x + 8 = 0 \Rightarrow 8 x (x - 8) - 1 (x - 8) = 0 \Rightarrow (x - 8) (8 x - 1) = 0 \Rightarrow x - 8 = 0 or 8 x - 1 = 0$
$\Rightarrow x = 8 or x = \frac{1}{8}$
∴ x = 8 (x is a natural number)

Hence, the required number is 8.

Page No 222:

Question 18:

Divide 57 into two parts whose product is 680.

Answer:

Let the two parts be x and (57 − x).

According to the given condition,
$x (57 - x) = 680 \Rightarrow 57 x - x^{2} = 680 \Rightarrow x^{2} - 57 x + 680 = 0 \Rightarrow x^{2} - 40 x - 17 x + 680 = 0$
$\Rightarrow x (x - 40) - 17 (x - 40) = 0 \Rightarrow (x - 40) (x - 17) = 0 \Rightarrow x - 40 = 0 or x - 17 = 0 \Rightarrow x = 40 or x = 17$

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.

Page No 222:

Question 19:

Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$ .

Answer:

Let the two parts be x and (27 − x).

According to the given condition,
$\frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20} \Rightarrow \frac{27 - x + x}{x (27 - x)} = \frac{3}{20} \Rightarrow \frac{27}{27 x - x^{2}} = \frac{3}{20} \Rightarrow 27 x - x^{2} = 180$
$\Rightarrow x^{2} - 27 x + 180 = 0 \Rightarrow x^{2} - 15 x - 12 x + 180 = 0 \Rightarrow x (x - 15) - 12 (x - 15) = 0 \Rightarrow (x - 12) (x - 15) = 0$
$\Rightarrow x - 12 = 0 or x - 15 = 0 \Rightarrow x = 12 or x = 15$

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.

Page No 222:

Question 20:

Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.

Answer:

$Let the larger and smaller parts be x and y, respectively . According to the question: x + y = 16 ... (i) 2 x^{2} = y^{2} + 164 ... (ii) From (i), we get: x = 16 - y ... (iii) From (ii) and (iii), we get: 2 {(16 - y)}^{2} = y^{2} + 164 \Rightarrow 2 (256 - 32 y + y^{2}) = y^{2} + 164 \Rightarrow 512 - 64 y + 2 y^{2} = y^{2} + 164 \Rightarrow y^{2} - 64 y + 348 = 0 \Rightarrow y^{2} - (58 + 6) y + 348 = 0 \Rightarrow y^{2} - 58 y - 6 y + 348 = 0 \Rightarrow y (y - 58) - 6 (y - 58) = 0 \Rightarrow (y - 58) (y - 6) = 0 \Rightarrow y - 58 = 0 or y - 6 = 0 \Rightarrow y = 6 (∵ y < 16) Putting the value of y in equation (iii), we get : x = 16 - 6 = 10 Hence, the two natural numbers are 6 and 10 .$

Page No 222:

Question 21:

Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

Answer:

$Let the two natural numbers be x and y . According to the question: x^{2} + y^{2} = 25 (x + y) . . . (i) x^{2} + y^{2} = 50 (x - y) . . . (ii) From (i) and (ii), we get: 25 (x + y) = 50 (x - y) \Rightarrow x + y = 2 (x - y) \Rightarrow x + y = 2 x - 2 y \Rightarrow y + 2 y = 2 x - x \Rightarrow 3 y = x . . . (iii) From (ii) and (iii), we get : {(3 y)}^{2} + y^{2} = 50 (3 y - y) \Rightarrow 9 y^{2} + y^{2} = 100 y \Rightarrow 10 y^{2} = 100 y \Rightarrow y = 10 From (iii), we have : 3 \times 10 = x \Rightarrow 30 = x Hence, the two natural numbers are 30 and 10 .$

Page No 222:

Question 22:

The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.

Answer:

$Let the greater number be x and the smaller number be y . According to the question: x^{2} - y^{2} = 45 . . . (i) y^{2} = 4 x . . . (ii) From (i) and (ii), we get: x^{2} - 4 x = 45 \Rightarrow x^{2} - 4 x - 45 = 0 \Rightarrow x^{2} - (9 - 5) x - 45 = 0 \Rightarrow x^{2} - 9 x + 5 x - 45 = 0 \Rightarrow x (x - 9) + 5 (x - 9) = 0 \Rightarrow (x - 9) (x + 5) = 0 \Rightarrow x - 9 = 0 or x + 5 = 0 \Rightarrow x = 9 or x = - 5 \Rightarrow x = 9 (∵ x is a natural number) Putting the value of x in equation (ii), we get : y^{2} = 4 \times 9 \Rightarrow y^{2} = 36 \Rightarrow y = 6 Hence, the two numbers are 9 and 6 .$

Page No 222:

Question 23:

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers. [CBSE 2010]

Answer:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,
$x^{2} + (x + 1) (x + 2) = 46 \Rightarrow x^{2} + x^{2} + 3 x + 2 = 46 \Rightarrow 2 x^{2} + 3 x - 44 = 0 \Rightarrow 2 x^{2} + 11 x - 8 x - 44 = 0$
$\Rightarrow x (2 x + 11) - 4 (2 x + 11) = 0 \Rightarrow (2 x + 11) (x - 4) = 0 \Rightarrow 2 x + 11 = 0 or x - 4 = 0 \Rightarrow x = - \frac{11}{2} or x = 4$
∴ x = 4 (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

Page No 222:

Question 24:

A two-digit number is 4 times the sum of its digits and twice the product of its digit. Find the number.

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ O riginal number = 10 y + x According to the question: 10 y + x = 4 (x + y) \Rightarrow 10 y + x = 4 x + 4 y \Rightarrow 3 x - 6 y = 0 \Rightarrow 3 x = 6 y \Rightarrow x = 2 y .... (i) Also, 10 y + x = 2 x y \Rightarrow 10 y + 2 y = 2 .2 y . y [From (i)] \Rightarrow 12 y = 4 y^{2} \Rightarrow y = 3 From (i), we get: x = 2 \times 3 = 6 ∴ Original number = 10 \times 3 + 6 = 36$

Page No 222:

Question 25:

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

Answer:

$Let the digits at units and tens places be x and y, respectively . ∴ x y =14 ⇒ y = \frac{14}{x} ... (i) According to the question: (10 y + x) + 45 = 10 x + y \Rightarrow 9 y - 9 x = - 45 \Rightarrow y - x = - 5 ... (ii) From (i) and (ii), we get: \frac{14}{x} - x = - 5 \Rightarrow \frac{14 - x^{2}}{x} = - 5 \Rightarrow 14 - x^{2} = - 5 x \Rightarrow x^{2} - 5 x - 14 = 0 \Rightarrow x^{2} - (7 - 2) x - 14 = 0 \Rightarrow x^{2} - 7 x + 2 x - 14 = 0 \Rightarrow x (x - 7) + 2 (x - 7) = 0 \Rightarrow (x - 7) (x + 2) = 0 \Rightarrow x - 7 = 0 or x +2=0 \Rightarrow x = 7 or x = - 2 \Rightarrow x = 7 (∵ the digit cannot be negative) Putting x =7 in equation (i), we get: y =2 ∴ Required number=10 \times 2+7=27$

Page No 222:

Question 26:

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2 \frac{9}{10}$ . Find the fraction.

Answer:

$Let the numerator be x . ∴ Denominator = x + 3 ∴ Original number = \frac{x}{x + 3} According to the question: \frac{x}{x + 3} + \frac{1}{(\frac{x}{x + 3})} = 2 \frac{9}{10} \Rightarrow \frac{x}{x + 3} + \frac{x + 3}{x} = \frac{29}{10} \Rightarrow \frac{x^{2} + {(x + 3)}^{2}}{x (x + 3)} = \frac{29}{10} \Rightarrow \frac{x^{2} + x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow \frac{2 x^{2} + 6 x + 9}{x^{2} + 3 x} = \frac{29}{10} \Rightarrow 29 x^{2} + 87 x = 20 x^{2} + 60 x + 90 \Rightarrow 9 x^{2} + 27 x - 90 = 0 \Rightarrow 9 (x^{2} + 3 x - 10) = 0 \Rightarrow x^{2} + 3 x - 10 = 0 \Rightarrow x^{2} + 5 x - 2 x - 10 = 0 \Rightarrow x (x + 5) - 2 (x + 5) = 0 \Rightarrow (x - 2) (x + 5) = 0 \Rightarrow x - 2 = 0 or x + 5 = 0 \Rightarrow x = 2 or x = - 5 (rejected) So, numerator = x = 2 denominator = x + 3 = 2 + 3 = 5 So, required fraction = \frac{2}{5}$

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Question 27:

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$ . Find the fraction. [CBSE 2012]

Answer:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = $\frac{x - 3}{x}$
If 1 is added to the denominator, then the new fraction obtained is $\frac{x - 3}{x + 1}$ .

According to the given condition,
$\frac{x - 3}{x + 1} = \frac{x - 3}{x} - \frac{1}{15} \Rightarrow \frac{x - 3}{x} - \frac{x - 3}{x + 1} = \frac{1}{15} \Rightarrow \frac{(x - 3) (x + 1) - x (x - 3)}{x (x + 1)} = \frac{1}{15} \Rightarrow \frac{x^{2} - 2 x - 3 - x^{2} + 3 x}{x^{2} + x} = \frac{1}{15}$
$\Rightarrow \frac{x - 3}{x^{2} + x} = \frac{1}{15} \Rightarrow x^{2} + x = 15 x - 45 \Rightarrow x^{2} - 14 x + 45 = 0 \Rightarrow x^{2} - 9 x - 5 x + 45 = 0$
$\Rightarrow x (x - 9) - 5 (x - 9) = 0 \Rightarrow (x - 5) (x - 9) = 0 \Rightarrow x - 5 = 0 or x - 9 = 0 \Rightarrow x = 5 or x = 9$

When x = 5,
$\frac{x - 3}{x} = \frac{5 - 3}{5} = \frac{2}{5}$

When x = 9,
$\frac{x - 3}{x} = \frac{9 - 3}{9} = \frac{6}{9} = \frac{2}{3}$ (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$ .

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Question 28:

The sum of a number and its reciprocal is $2 \frac{1}{30}$ . Find the number.

Answer:

Let the required number be x.

According to the given condition,
$x + \frac{1}{x} = 2 \frac{1}{30} \Rightarrow \frac{x^{2} + 1}{x} = \frac{61}{30} \Rightarrow 30 x^{2} + 30 = 61 x \Rightarrow 30 x^{2} - 61 x + 30 = 0$
$\Rightarrow 30 x^{2} - 36 x - 25 x + 30 = 0 \Rightarrow 6 x (5 x - 6) - 5 (5 x - 6) = 0 \Rightarrow (5 x - 6) (6 x - 5) = 0 \Rightarrow 5 x - 6 = 0 or 6 x - 5 = 0$
$\Rightarrow x = \frac{6}{5} or x = \frac{5}{6}$
Hence, the required number is $\frac{5}{6}$ or $\frac{6}{5}$ .

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Question 29:

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.

Answer:

$Let there be x rows. Then, the number of students in each row will also be x . ∴ T otal number of students = (x^{2} + 24) According to the question: {(x + 1)}^{2} - 25 = x^{2} + 24 \Rightarrow x^{2} + 2 x + 1 - 25 - x^{2} - 24 = 0 \Rightarrow 2 x - 48 = 0 \Rightarrow 2 x = 48 \Rightarrow x = 24 ∴ {Total number of students = 24}^{2} + 24 = 576 + 24 = 600$

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Question 30:

300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of student.

Answer:

$\begin{array}{l} Let the total number of students be x . \end{array} According to the question : \frac{300}{x} - \frac{300}{x + 10} = 1 \Rightarrow \frac{300 (x + 10) - 300 x}{x (x + 10)} = 1 \Rightarrow \frac{300 x + 3000 - 300 x}{x^{2} + 10 x} = 1 \Rightarrow 3000 = x^{2} + 10 x \Rightarrow x^{2} + 10 x - 3000 = 0 \Rightarrow x^{2} + (60 - 50) x - 3000 = 0 \Rightarrow x^{2} + 60 x - 50 x - 3000 = 0 \Rightarrow x (x + 60) - 50 (x + 60) = 0 \Rightarrow (x + 60) (x - 50) = 0 \Rightarrow x = 50 or x = - 60 x cannot be negative; therefore, the total number of students is 50.$

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Question 31:

In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.

Answer:

$\begin{array}{l} Let the marks of Kamal in mathematics and english be x and y, respectively. \end{array} According to the question : x + y = 40 . . . (i) Also, (x + 3) (y - 4) = 360 \Rightarrow (x + 3) (40 - x - 4) = 360 [F rom (i)] \Rightarrow (x + 3) (36 - x) = 360 \Rightarrow 36 x - x^{2} + 108 - 3 x = 360 \Rightarrow 33 x - x^{2} - 252 = 0 \Rightarrow - x^{2} + 33 x - 252 = 0 \Rightarrow x^{2} - 33 x + 252 = 0 \Rightarrow x^{2} - (21 + 12) x + 252 = 0 \Rightarrow x^{2} - 21 x - 12 x + 252 = 0 \Rightarrow x (x - 21) - 12 (x - 21) = 0 \Rightarrow (x - 21) (x - 12) = 0 \Rightarrow x = 21 or x = 12 If x = 21, y = 40 - 21 = 19 Thus, Kamal scored 21 and 19 m arks in mathematics and english, respectively. If x = 12, y = 40 - 12 = 28 Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.$

Page No 223:

Question 32:

Some students planned a picnic. The total budget for food was ₹2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food? [CBSE 2010]

Answer:

Let x be the number of students who planned a picnic.

∴ Original cost of food for each member = ₹ $\frac{2000}{x}$

Five students failed to attend the picnic. So, (x − 5) students attended the picnic.

∴ New cost of food for each member = ₹ $\frac{2000}{(x - 5)}$

According to the given condition,
₹ $\frac{2000}{x - 5}$ − ₹ $\frac{2000}{x}$ = ₹20
$\Rightarrow \frac{2000 x - 2000 x + 10000}{x (x - 5)} = 20 \Rightarrow \frac{10000}{x^{2} - 5 x} = 20 \Rightarrow x^{2} - 5 x = 500 \Rightarrow x^{2} - 5 x - 500 = 0$
$\Rightarrow x^{2} - 25 x + 20 x - 500 = 0 \Rightarrow x (x - 25) + 20 (x - 25) = 0 \Rightarrow (x - 25) (x + 20) = 0 \Rightarrow x - 25 = 0 or x + 20 = 0$
$\Rightarrow x = 25 or x = - 20$
∴ x = 25 (Number of students cannot be negative)

Number of students who attended the picnic = x − 5 = 25 − 5 = 20

Amount paid by each student for the food = ₹ $\frac{2000}{(25 - 5)}$ = ₹ $\frac{2000}{20}$ = ₹100

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Question 33:

If the price of a book is reduced by ₹5, a person can buy 4 more books for ₹600. Find the original price of the book.

Answer:

Let the original price of the book be ₹x.

∴ Number of books bought at original price for ₹600 = $\frac{600}{x}$

If the price of a book is reduced by ₹5, then the new price of the book is ₹(x − 5).

∴ Number of books bought at reduced price for ₹600 = $\frac{600}{x - 5}$

According to the given condition,
$\frac{600}{x - 5} - \frac{600}{x} = 4 \Rightarrow \frac{600 x - 600 x + 3000}{x (x - 5)} = 4 \Rightarrow \frac{3000}{x^{2} - 5 x} = 4 \Rightarrow x^{2} - 5 x = 750$
$\Rightarrow x^{2} - 5 x - 750 = 0 \Rightarrow x^{2} - 30 x + 25 x - 750 = 0 \Rightarrow x (x - 30) + 25 (x - 30) = 0 \Rightarrow (x - 30) (x + 25) = 0$
$\Rightarrow x - 30 = 0 or x + 25 = 0 \Rightarrow x = 30 or x = - 25$
∴ x = 30 (Price cannot be negative)

Hence, the original price of the book is ₹30.

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Question 34:

A person on tour has ₹10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by ₹90. Find the original duration of the tour.

Answer:

Let the original duration of the tour be x days.

∴ Original daily expenses = ₹ $\frac{10, 800}{x}$

If he extends his tour by 4 days, then his new daily expenses = ₹ $\frac{10, 800}{x + 4}$

According to the given condition,

₹ $\frac{10, 800}{x}$ − ₹ $\frac{10, 800}{x + 4}$ = ₹90

$\Rightarrow \frac{10800 x + 43200 - 10800 x}{x (x + 4)} = 90 \Rightarrow \frac{43200}{x^{2} + 4 x} = 90 \Rightarrow x^{2} + 4 x = 480 \Rightarrow x^{2} + 4 x - 480 = 0$
$\Rightarrow x^{2} + 24 x - 20 x - 480 = 0 \Rightarrow x (x + 24) - 20 (x + 24) = 0 \Rightarrow (x + 24) (x - 20) = 0 \Rightarrow x + 24 = 0 or x - 20 = 0$
$\Rightarrow x = - 24 or x = 20$
∴ x = 20 (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

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Question 35:

In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained him in the two subjects separately. [CBSE 2008]

Answer:

Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.

According to the given condition,
$(x + 3) (28 - x - 4) = 180 \Rightarrow (x + 3) (24 - x) = 180 \Rightarrow - x^{2} + 21 x + 72 = 180 \Rightarrow x^{2} - 21 x + 108 = 0$
$\Rightarrow x^{2} - 12 x - 9 x + 108 = 0 \Rightarrow x (x - 12) - 9 (x - 12) = 0 \Rightarrow (x - 12) (x - 9) = 0 \Rightarrow x - 12 = 0 or x - 9 = 0$
$\Rightarrow x = 12 or x = 9$

When x = 12,
28 − x = 28 − 12 = 16

When x = 9,
28 − x = 28 − 9 = 19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

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Question 36:

A man buys a number of pens for ₹ 180. If he had bought 3 more pens for the same amount, each pen would have cost him ₹ 3 less. How many pens did he buy?

Answer:

Let number of pens bought = x
Let the price per pen = Rs y
It is given that the price of x pens is Rs 180
$\Rightarrow x y = 180 . . . (i) \Rightarrow x = \frac{180}{y}$
Also, given that if 3 more pens are purchased for the same amount, the price per pen gets reduced by Rs 3
$\Rightarrow (y - 3) (x + 3) = 180 . . . (i i)$
Put the value of x in (ii) we get
$(\frac{180}{x} - 3) (x + 3) = 180 \Rightarrow (\frac{180 - 3 x}{x}) (x + 3) = 180 \Rightarrow (180 - 3 x) (x + 3) = 180 x \Rightarrow 180 x + 540 - 3 x^{2} - 9 x = 180 x \Rightarrow 3 x^{2} + 9 x - 540 = 0 \Rightarrow x^{2} + 3 x - 180 = 0 \Rightarrow x^{2} + 15 x - 12 x - 180 = 0 \Rightarrow x (x + 15) - 12 (x + 15) = 0 \Rightarrow (x + 15) (x - 12) = 0 \Rightarrow x + 15 = 0 or x - 12 = 0 \Rightarrow x = - 15 or x = 12$
We ignore the negative value.
$\Rightarrow x = 12$
Therefore, number of pens is 12.

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Question 37:

A dealer sells an article for ₹75 and gains as much percent as the cost price of the article. Find the cost price of the article.
[CBSE 2011]

Answer:

Let the cost price of the article be ₹x.

∴ Gain percent = x%

According to the given condition,
₹x + ₹ $(\frac{x}{100} \times x)$ = ₹75 (Cost price + Gain = Selling price)
$\Rightarrow \frac{100 x + x^{2}}{100} = 75 \Rightarrow x^{2} + 100 x = 7500 \Rightarrow x^{2} + 100 x - 7500 = 0 \Rightarrow x^{2} + 150 x - 50 x - 7500 = 0$
$\Rightarrow x (x + 150) - 50 (x + 150) = 0 \Rightarrow (x - 50) (x + 150) = 0 \Rightarrow x - 50 = 0 or x + 150 = 0 \Rightarrow x = 50 or x = - 150$
∴ x = 50 (Cost price cannot be negative)

Hence, the cost price of the article is ₹50.

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Question 38:

(i) One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.
(ii) A man is $3 \frac{1}{2}$ times as old as his son. If the sum of the squares of their ages is 1325, find the ages of the father and the son. (CBSE 2017]

Answer:

(i)
Let the present age of the son be x years.

∴ Present age of the man = x² years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x² − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$∴ x^{2} - 1 = 8 (x - 1) \Rightarrow x^{2} - 1 = 8 x - 8 \Rightarrow x^{2} - 8 x + 7 = 0 \Rightarrow x^{2} - 7 x - x + 7 = 0$
$\Rightarrow x (x - 7) - 1 (x - 7) = 0 \Rightarrow (x - 1) (x - 7) = 0 \Rightarrow x - 1 = 0 or x - 7 = 0 \Rightarrow x = 1 or x = 7$
∴ x = 7 (Man's age cannot be 1 year)

Present age of the son = 7 years

Present age of the man = 7² years = 49 years

(ii)
Let the age of man be m and the age of son be s
It is given that man is $3 \frac{1}{2}$ times as old as his son.
$\Rightarrow m = 3 \frac{1}{2} s \Rightarrow m = \frac{7}{2} s . . . (i)$
Also given that
$m^{2} + s^{2} = 1325 . . . . . (ii)$
Put value of (i) in (ii), we get
${(\frac{7}{2} s)}^{2} + s^{2} = 1325 \frac{\Rightarrow 49 s^{2} + 4 s^{2}}{4} = 1325 53 s^{2} = 5300 \Rightarrow s^{2} = 100 \Rightarrow s = \pm 10$
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
$m = \frac{7}{2} s \Rightarrow m = \frac{7}{2} \times 10 \Rightarrow m = 35$
So, age of man = 35 years
Age of son = 10 years

Page No 224:

Question 39:

The sum of the reciprocals of Meena's ages (in years) 3 years ago and 5 years hence is $\frac{1}{3}$ . Find her present age.

Answer:

Let the present age of Meena be x years.

Meena's age 3 years ago = (x − 3) years

Meena's age 5 years hence = (x + 5) years

According to the given condition,
$\frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \Rightarrow \frac{x + 5 + x - 3}{(x - 3) (x + 5)} = \frac{1}{3} \Rightarrow \frac{2 x + 2}{x^{2} + 2 x - 15} = \frac{1}{3} \Rightarrow x^{2} + 2 x - 15 = 6 x + 6$
$\Rightarrow x^{2} - 4 x - 21 = 0 \Rightarrow x^{2} - 7 x + 3 x - 21 = 0 \Rightarrow x (x - 7) + 3 (x - 7) = 0 \Rightarrow (x - 7) (x + 3) = 0$
$\Rightarrow x - 7 = 0 or x + 3 = 0 \Rightarrow x = 7 or x = - 3$
∴ x = 7 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

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Question 40:

The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.

Answer:

$Let the present ages of the boy and his brother be x years and (25 - x) years . According to the question : x (25 - x) = 126 \Rightarrow 25 x - x^{2} = 126 \Rightarrow x^{2} - (18 + 7) x + 126 = 0 \Rightarrow x^{2} - 18 x - 7 x + 126 = 0 \Rightarrow x (x - 18) - 7 (x - 18) = 0 \Rightarrow (x - 18) (x - 7) = 0 \Rightarrow x - 18 = 0 or x - 7 = 0 \Rightarrow x = 18 or x = 7 \Rightarrow x = 18 (∵ P resent age of the boy cannot be less than his brother) If x = 18, we have : Present ages of the boy = 18 years Present age of his brother = (25 - 18) years = 7 years Thus, the present ages of the boy and his brother are 18 years and 7 years, respectively.$

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Question 41:

The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age.

Answer:

$Let the present age of Meena be x years . According to the question : (x - 5) (x + 8) = 30 \Rightarrow x^{2} + 3 x - 40 = 30 \Rightarrow x^{2} + 3 x - 70 = 0 \Rightarrow x^{2} + (10 - 7) x - 70 = 0 \Rightarrow x^{2} + 10 x - 7 x - 70 = 0 \Rightarrow x (x + 10) - 7 (x + 10) = 0 \Rightarrow (x + 10) (x - 7) = 0 \Rightarrow x + 10 = 0 or x - 7 = 0 \Rightarrow x = - 10 or x = 7 \Rightarrow x = 7 (∵ Age cannot be negative) Thus, the present age of Meena is 7 years .$

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Question 42:

Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.

Answer:

Let son's age 2 years ago be x years. Then,

Man's age 2 years ago = 3x²years

∴ Son's present age = (x + 2) years

Man's present age = (3x² + 2) years

In three years time,

Son's age = (x + 2 + 3) years = (x + 5) years

Man's age = (3x² + 2 + 3) years = (3x² + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3x² + 5 = 4(x + 5)

$\Rightarrow 3 x^{2} + 5 = 4 x + 20 \Rightarrow 3 x^{2} - 4 x - 15 = 0 \Rightarrow 3 x^{2} - 9 x + 5 x - 15 = 0 \Rightarrow 3 x (x - 3) + 5 (x - 3) = 0$
$\Rightarrow (x - 3) (3 x + 5) = 0 \Rightarrow x - 3 = 0 or 3 x + 5 = 0 \Rightarrow x = 3 or x = - \frac{5}{3}$
∴ x = 3 (Age cannot be negative)

Son's present age = (x + 2) years = (3 + 2) years = 5 years

Man's present age = (3x² + 2) years = (3× 9 + 2) years = 29 years

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Question 43:

A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck. [CBSE 2015]

Answer:

Let the first speed of the truck be x km/h.

∴ Time taken to cover 150 km = $\frac{150}{x}$ h $(Time = \frac{Distance}{Speed})$

New speed of the truck = (x + 20) km/h

∴ Time taken to cover 200 km = $\frac{200}{x + 20}$ h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
$∴ \frac{150}{x} + \frac{200}{x + 20} = 5 \Rightarrow \frac{150 x + 3000 + 200 x}{x (x + 20)} = 5 \Rightarrow 350 x + 3000 = 5 (x^{2} + 20 x) \Rightarrow 350 x + 3000 = 5 x^{2} + 100 x$
$\Rightarrow 5 x^{2} - 250 x - 3000 = 0 \Rightarrow x^{2} - 50 x - 600 = 0 \Rightarrow x^{2} - 60 x + 10 x - 600 = 0 \Rightarrow x (x - 60) + 10 (x - 60) = 0$
$\Rightarrow (x - 60) (x + 10) = 0 \Rightarrow x - 60 = 0 or x + 10 = 0 \Rightarrow x = 60 or x = - 10$
∴ x = 60 (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.

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Question 44:

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes. To reach the destination, 1500 km away, in time, the pilot increased the speed by 100 km/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?
[CBSE 2013]

Answer:

Let the original speed of the plane be x km/h.

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = $\frac{1500}{x}$ h $(Time = \frac{Distance}{Speed})$

Time taken to reach the destination at actual speed = $\frac{1500}{x + 100}$ h

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

$∴ \frac{1500}{x} = \frac{1500}{x + 100} + \frac{1}{2} (30 \min = \frac{30}{60} h = \frac{1}{2} h) \Rightarrow \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} \Rightarrow \frac{1500 x + 150000 - 1500 x}{x (x + 100)} = \frac{1}{2} \Rightarrow \frac{150000}{x^{2} + 100 x} = \frac{1}{2}$
$\Rightarrow x^{2} + 100 x = 300000 \Rightarrow x^{2} + 100 x - 300000 = 0 \Rightarrow x^{2} + 600 x - 500 x - 300000 = 0 \Rightarrow x (x + 600) - 500 (x + 600) = 0$
$\Rightarrow (x + 600) (x - 500) = 0 \Rightarrow x + 600 = 0 or x - 500 = 0 \Rightarrow x = - 600 or x = 500$
∴ x = 500 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

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Question 45:

A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train. [CBSE 2013C]

Answer:

Let the usual speed of the train be x km/h.

∴ Reduced speed of the train = (x − 8) km/h

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = $\frac{480}{x}$ h $(Time = \frac{Distance}{Speed})$

Time taken by the train to cover the distance at reduced speed = $\frac{480}{x - 8}$ h

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

$∴ \frac{480}{x - 8} = \frac{480}{x} + 3 \Rightarrow \frac{480}{x - 8} - \frac{480}{x} = 3 \Rightarrow \frac{480 x - 480 x + 3840}{x (x - 8)} = 3 \Rightarrow \frac{3840}{x^{2} - 8 x} = 3$
$\Rightarrow x^{2} - 8 x = 1280 \Rightarrow x^{2} - 8 x - 1280 = 0 \Rightarrow x^{2} - 40 x + 32 x - 1280 = 0 \Rightarrow x (x - 40) + 32 (x - 40) = 0$
$\Rightarrow (x - 40) (x + 32) = 0 \Rightarrow x - 40 = 0 or x + 32 = 0 \Rightarrow x = 40 or x = - 32$
∴ x = 40 (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.

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Question 46:

A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed? [CBSE 2015]

Answer:

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = $\frac{54}{x}$ h $(Time = \frac{Distance}{Speed})$

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = $\frac{63}{x + 6}$ h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
$∴ \frac{54}{x} + \frac{63}{x + 6} = 3 \Rightarrow \frac{54 x + 324 + 63 x}{x (x + 6)} = 3 \Rightarrow 117 x + 324 = 3 (x^{2} + 6 x) \Rightarrow 117 x + 324 = 3 x^{2} + 18 x$
$\Rightarrow 3 x^{2} - 99 x - 324 = 0 \Rightarrow x^{2} - 33 x - 108 = 0 \Rightarrow x^{2} - 36 x + 3 x - 108 = 0 \Rightarrow x (x - 36) + 3 (x - 36) = 0$
$\Rightarrow (x - 36) (x + 3) = 0 \Rightarrow x - 36 = 0 or x + 3 = 0 \Rightarrow x = 36 or x = - 3$
∴ x = 36 (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.

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Question 47:

A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train. [CBSE 2013]

Answer:

Distance covered by the train = 180 km
We know that distance covered $(d) = speed (s) \times time (t)$
$\Rightarrow s \times t = 180 \Rightarrow s = \frac{180}{t} . . . (i)$
Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour.
$\Rightarrow (s + 9) (t - 1) = 180 . . . (ii)$
Put the value of (i) in (ii), we get
$(\frac{180}{t} + 9) (t - 1) = 180 (180 + 9 t) (t - 1) = 180 t 180 t - 180 + 9 t^{2} - 9 t = 180 t 9 t^{2} - 9 t - 180 = 0 t^{2} - t - 20 = 0 t^{2} - 5 t + 4 t - 20 = 0 t (t - 5) + 4 (t - 5) = 0 (t + 4) (t - 5) = 0 (t + 4) = 0 or (t - 5) = 0 t = - 4 or t = 5$
Ignore the negative value
So, time taken = 5 hours
From (i)
$s = \frac{180}{t} = \frac{180}{5} = 36$
Hence, the speed is 36 km/h.

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Question 52:

The speed of a boat in still water is 15 km/hr. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE 2017]

Answer:

Let, speed of stream be x km/h
Speed of boat = 15 km/h
Distance from each side = 30 km
We know that time taken $= \frac{distance covered}{Speed}$
Total speed of the boat while going upstream $= 15 - x$ km/h
Time taken to go upstream $= \frac{30}{15 - x}$ hrs
Total speed of boat while going downstream $= 15 + x$ km/h
Time taken to go downstream $= \frac{30}{15 + x}$ hrs
Total time of the journey $= 4 \frac{1}{2}$ hrs = 4.5 hrs
$\Rightarrow \frac{30}{15 - x} + \frac{30}{15 + x} = 4.5 \Rightarrow \frac{30 [(15 + x) + (15 - x)]}{(15)^{2} - x^{2}} = 4.5 \Rightarrow 30 [15 + x + 15 - x] = 4.5 [(15)^{2} - x^{2}] \Rightarrow 30 [30] = 4.5 [225 - x^{2}] \Rightarrow \frac{30 \times 30}{4.5} = 225 - x^{2} \Rightarrow 225 - x^{2} = 200 \Rightarrow x^{2} = 25 \Rightarrow x = \pm 5$
Ignore the negative value.
So, the speed of the stream = x = 5 km/h

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Question 48:

A train covers a distance of 300 km at a uniform speed. If the speed of the train is increased by 5 km/hour, it takes 2 hours less in the journey. Find the original speed of the train.

Answer:

Distance covered by the train = 300 km
We know that distance covered $(d) = speed (s) \times time (t)$
$\Rightarrow s \times t = 300 \Rightarrow t = \frac{300}{s} . . . (i)$
Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours.
$\Rightarrow (s + 5) (t - 2) = 300 . . . . . (ii)$
Put the value of (i) in (ii), we get
$(s + 5) (\frac{300}{s} - 2) = 300 (s + 5) (\frac{300 - 2 s}{s}) = 300 (s + 5) (300 - 2 s) = 300 s 300 s - 2 s^{2} + 1500 - 10 s = 300 s - 2 s^{2} + 1500 - 10 s = 0 - s^{2} + 750 - 5 s = 0 s^{2} + 5 s - 750 = 0 s^{2} + 30 s - 25 s - 750 = 0 s (s + 30) - 25 (s + 30) = 0 (s - 25) (s + 30) = 0 (s - 25) = 0 or (s + 30) = 0 s = 25 or s = - 30$
Ignore the negative value
Therefore, the original speed = 25 km/h

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Question 49:

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

Answer:

$\begin{array}{l} Let the usual speed be x km/hr . \end{array} According to the question : \frac{300}{x} - \frac{300}{(x + 5)} = 2 \Rightarrow \frac{300 (x + 5) - 300 x}{x (x + 5)} = 2 \Rightarrow \frac{300 x + 1500 - 300 x}{x^{2} + 5 x} = 2 \Rightarrow 1500 = 2 (x^{2} + 5 x) \Rightarrow 1500 = 2 x^{2} + 10 x \Rightarrow x^{2} + 5 x - 750 = 0 \Rightarrow x^{2} + (30 - 25) x - 750 = 0 \Rightarrow x^{2} + 30 x - 25 x - 750 = 0 \Rightarrow x (x + 30) - 25 (x + 30) = 0 \Rightarrow (x + 30) (x - 25) = 0 \Rightarrow x = - 30 or x = 25 T he usual speed cannot be negative; therefore, the speed is 25 km/hr.$

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Question 50:

The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by 20 km/hr.

Answer:

$\begin{array}{l} Let the speed of the Deccan Queen be x km/hr . \end{array} According to the question : Speed of another train = (x - 20) km / h ∴ \frac{192}{x - 20} - \frac{192}{x} = \frac{48}{60} \Rightarrow \frac{4}{x - 20} - \frac{4}{x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 (x - 20)}{(x - 20) x} = \frac{1}{60} \Rightarrow \frac{4 x - 4 x + 80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow \frac{80}{x^{2} - 20 x} = \frac{1}{60} \Rightarrow x^{2} - 20 x = 4800 \Rightarrow x^{2} - 20 x - 4800 = 0 \Rightarrow x^{2} - (80 - 60) x - 4800 = 0 \Rightarrow x^{2} - 80 x + 60 x - 4800 = 0 \Rightarrow x (x - 80) + 60 (x - 80) = 0 \Rightarrow (x - 80) (x + 60) = 0 \Rightarrow x = 80 or x = - 60 The value of x cannot be negative; therefore, the original speed of Deccan Queen is 80 km/hr.$

Page No 225:

Question 51:

A motor boat whose speed in still water is 18 km/hr, takes 1 hour more to go 24 km upstream than o return to the same spot. Find the speed of the stream.

Answer:

$\begin{array}{l} Let the speed of the stream be x km/hr. \end{array} Given : Speed of the boat = 18 km / hr ∴ Speed downstream = (18 + x) km/hr S peed upstream = (18 - x) km/hr ∴ \frac{24}{(18 - x)} - \frac{24}{(18 + x)} = 1 \Rightarrow \frac{1}{(18 - x)} - \frac{1}{(18 + x)} = \frac{1}{24} \Rightarrow \frac{18 + x - 18 + x}{(18 - x) (18 + x)} = \frac{1}{24} \Rightarrow \frac{2 x}{18^{2} - x^{2}} = \frac{1}{24} \Rightarrow 324 - x^{2} = 48 x \Rightarrow 324 - x^{2} - 48 x = 0 \Rightarrow x^{2} + 48 x - 324 = 0 \Rightarrow x^{2} + (54 - 6) x - 324 = 0 \Rightarrow x^{2} + 54 x - 6 x - 324 = 0 \Rightarrow x (x + 54) - 6 (x + 54) = 0 \Rightarrow (x + 54) (x - 6) = 0 \Rightarrow x = - 54 or x = 6 The value of x cannot be negative; thereore, the speed of the stream is 6 km/hr.$

Page No 225:

Question 53:

A motorboat whose speed is 9 km/hr in still water, goes 15 km downstream and comes back in a total time of 3 hours 45 minutes. Find the speed of the stream.

Answer:

$\begin{array}{l} Let the speed of the stream be x km / hr . \end{array} ∴ D ownstream speed= (9 + x) km/hr Upstream speed= (9 - x) km/hr Distance covered downstream = Distance covered upstream = 15 km Total time taken = 3 hours 45 minutes = (3 + \frac{45}{60}) minutes = \frac{225}{60} minutes = \frac{15}{4} minutes ∴ \frac{15}{(9 + x)} + \frac{15}{(9 - x)} = \frac{15}{4} \Rightarrow \frac{1}{(9 + x)} + \frac{1}{(9 - x)} = \frac{1}{4} \Rightarrow \frac{9 - x + 9 + x}{(9 + x) (9 - x)} = \frac{1}{4} \Rightarrow \frac{18}{9^{2} - x^{2}} = \frac{1}{4} \Rightarrow \frac{18}{81 - x^{2}} = \frac{1}{4} \Rightarrow 81 - x^{2} = 72 \Rightarrow 81 - x^{2} - 72 = 0 \Rightarrow - x^{2} + 9 = 0 \Rightarrow x^{2} = 9 \Rightarrow x = 3 or x = - 3 The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.$

Page No 225:

Question 54:

A takes 10 days than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

Answer:

$\begin{array}{l} Let B takes x days to complete the work. \end{array} Therefore, A will take (x - 10) days. ∴ \frac{1}{x} + \frac{1}{(x - 10)} = \frac{1}{12} \Rightarrow \frac{(x - 10) + x}{x (x - 10)} = \frac{1}{12} \Rightarrow \frac{2 x - 10}{x^{2} - 10 x} = \frac{1}{12} \Rightarrow x^{2} - 10 x = 12 (2 x - 10) \Rightarrow x^{2} - 10 x = 24 x - 120 \Rightarrow x^{2} - 34 x + 120 = 0 \Rightarrow x^{2} - (30 + 4) x + 120 = 0 \Rightarrow x^{2} - 30 x - 4 x + 120 = 0 \Rightarrow x (x - 30) - 4 (x - 30) = 0 \Rightarrow (x - 30) (x - 4) = 0 \Rightarrow x = 30 or x = 4 Number of days to complete the work by B cannot be less than that by A; therefore, we get: x = 30 Thus, B completes the work in 30 days .$

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Question 55:

Two taps running together can fill a tank in $3 \frac{1}{13}$ hours. If one tap takes 3 hours more than the other to fill the tank then how much time will each tap take to fill the tank?

Answer:

$Let one tap fills the tank in x hours. Therefore, the o ther tap will fill the tank in (x + 3) hours. Time taken by both taps, running together, to fill the tank= 3 \frac{1}{13} hours = \frac{40}{13} hours Part filled by one tap in 1 hour = \frac{1}{x} Part filled by the other tap in 1 hour = \frac{1}{x + 3} Part filled by both taps, running together, in 1 hour = \frac{1}{x} + \frac{1}{x + 3} ∴ \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{\frac{40}{13}} \Rightarrow \frac{(x + 3) + x}{x (x + 3)} = \frac{13}{40} \Rightarrow \frac{2 x + 3}{x^{2} + 3 x} = \frac{13}{40} \Rightarrow 13 x^{2} + 39 x = 80 x + 120 \Rightarrow 13 x^{2} - 41 x - 120 = 0 \Rightarrow 13 x^{2} - (65 - 24) x - 120 = 0 \Rightarrow 13 x^{2} - 65 x + 24 x - 120 = 0 \Rightarrow 13 x (x - 5) + 24 (x - 5) = 0 \Rightarrow (x - 5) (13 x + 24) = 0 \Rightarrow x - 5 = 0 or 13 x + 24 = 0 ⇒ x = 5 or x = \frac{- 24}{13} \Rightarrow x = 5 (∵ time cannot be a negative fraction) Thus, o ne pipe will take 5 hour and the other will take {(5 + 3) = 8} hour s to fill the tank.$

Page No 225:

Question 56:

Two pipes running together can fill a tank in $11 \frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately. [CBSE 2010]

Answer:

Let the time taken by one pipe to fill the tank be x minutes.

∴ Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

∴ Volume of the tank filled by one pipe in 1 minute = $\frac{V}{x}$

⇒ Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes = $\frac{V}{x} \times 11 \frac{1}{9} = \frac{V}{x} \times \frac{100}{9}$

Similarly,

Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes = $\frac{V}{(x + 5)} \times 11 \frac{1}{9} = \frac{V}{(x + 5)} \times \frac{100}{9}$

Now,

Volume of the tank filled by one pipe in $11 \frac{1}{9}$ minutes + Volume of the tank filled by the other pipe in $11 \frac{1}{9}$ minutes = V
$∴ V (\frac{1}{x} + \frac{1}{x + 5}) \times \frac{100}{9} = V \Rightarrow \frac{1}{x} + \frac{1}{x + 5} = \frac{9}{100} \Rightarrow \frac{x + 5 + x}{x (x + 5)} = \frac{9}{100} \Rightarrow \frac{2 x + 5}{x^{2} + 5 x} = \frac{9}{100}$
$\Rightarrow 200 x + 500 = 9 x^{2} + 45 x \Rightarrow 9 x^{2} - 155 x - 500 = 0 \Rightarrow 9 x^{2} - 180 x + 25 x - 500 = 0 \Rightarrow 9 x (x - 20) + 25 (x - 20) = 0$
$\Rightarrow (x - 20) (9 x + 25) = 0 \Rightarrow x - 20 = 0 or 9 x + 25 = 0 \Rightarrow x = 20 or x = - \frac{25}{9}$
∴ x = 20 (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min

Page No 225:

Question 57:

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. [CBSE 2011]

Answer:

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = $\frac{V}{x}$

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = $\frac{V}{x} \times 6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = $\frac{V}{(x - 9)} \times 6$

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
$∴ V (\frac{1}{x} + \frac{1}{x - 9}) \times 6 = V \Rightarrow \frac{1}{x} + \frac{1}{x - 9} = \frac{1}{6} \Rightarrow \frac{x - 9 + x}{x (x - 9)} = \frac{1}{6} \Rightarrow \frac{2 x - 9}{x^{2} - 9 x} = \frac{1}{6}$
$\Rightarrow 12 x - 54 = x^{2} - 9 x \Rightarrow x^{2} - 21 x + 54 = 0 \Rightarrow x^{2} - 18 x - 3 x + 54 = 0 \Rightarrow x (x - 18) - 3 (x - 18) = 0$
$\Rightarrow (x - 18) (x - 3) = 0 \Rightarrow x - 18 = 0 or x - 3 = 0 \Rightarrow x = 18 or x = 3$

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

∴ x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

Page No 225:

Question 58:

The length of a rectangle is twice its breadth and its area is 288 cm². Find the dimensions of the rectangle.

Answer:

$Let the length and breadth of the rectangle be 2 x m and x m, respectively. According to the question : 2 x \times x = 288 \Rightarrow 2 x^{2} = 288 \Rightarrow x^{2} = 144 \Rightarrow x = 12 or x = - 12 \Rightarrow x = 12 (∵ x cannot be negative) ∴ L ength = 2 \times 12 = 24 m Breadth = 12 m$

Page No 226:

Question 59:

The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.

Answer:

$Let the length and breadth of the rectangle be 3 x m and x m, respectively. According to the question : 3 x \times x = 147 \Rightarrow 3 x^{2} = 147 \Rightarrow x^{2} = 49 \Rightarrow x = 7 or x = - 7 \Rightarrow x = 7 (∵ x cannot be negative) ∴ Length = 3 \times 7 = 21m Breadth = 7 m$

Page No 226:

Question 60:

The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.

Answer:

$\begin{array}{l} Let the breadth of the rectangular hall be x metre. \end{array} Therefore, the length of the rectangular hall will be (x + 3) metre. According to the question : x (x + 3) = 238 \Rightarrow x^{2} + 3 x = 238 \Rightarrow x^{2} + 3 x - 238 = 0 \Rightarrow x^{2} + (17 - 14) x - 238 = 0 \Rightarrow x^{2} + 17 x - 14 x - 238 = 0 \Rightarrow x (x + 17) - 14 (x + 17) = 0 \Rightarrow (x + 17) (x - 14) = 0 \Rightarrow x = - 17 or x = 14 But the value x cannot be negative. Therefore, the breadth of the hall is 14 metre and the length is 17 metre.$

Page No 226:

Question 61:

The perimeter of a rectangular plot is 62 m and its area is 228 sq metres. Find the dimensions of the plot.

Answer:

$Let the length and breadth of the rectangular plot be x and y meter, respectively . Therefore, we have : Perimeter = 2 (x + y) = 62 . . . (i) and A rea = x y = 228 \Rightarrow y = \frac{228}{x} Putting the value of y in (i), we get ⇒ 2 (x + \frac{228}{x}) = 62 \Rightarrow x + \frac{228}{x} = 31 \Rightarrow \frac{x^{2} + 228}{x} = 31 \Rightarrow x^{2} + 228 = 31 x \Rightarrow x^{2} - 31 x + 228 = 0 \Rightarrow x^{2} - (19 + 12) x + 228 = 0 \Rightarrow x^{2} - 19 x - 12 x + 228 = 0 \Rightarrow x (x - 19) - 12 (x - 19) = 0 \Rightarrow (x - 19) (x - 12) = 0 \Rightarrow x = 19 or x = 12 If x = 19 m, y = \frac{228}{19} = 12 m Therefore, the length and breadth of the plot are 19 m and 12 m, respectively .$

Page No 226:

Question 62:

A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m². Find the width of the path.

Answer:

$Let the width of the path be x m. ∴ Length of the field including the path = 16 + x + x = 16 + 2 x Breadth of the field including the path = 10 + x + x = 10 + 2 x Now, (Area of the field including path) - (A rea of the field excluding path) = Area of the path \Rightarrow (16 + 2 x) (10 + 2 x) - (16 \times 10) = 120 \Rightarrow 160 + 32 x + 20 x + 4 x^{2} - 160 = 120 \Rightarrow 4 x^{2} + 52 x - 120 = 0 \Rightarrow x^{2} + 13 x - 30 = 0 \Rightarrow x^{2} + (15 - 2) x + 30 = 0 \Rightarrow x^{2} + 15 x - 2 x + 30 = 0 \Rightarrow x (x + 15) - 2 (x + 15) = 0 \Rightarrow (x - 2) (x + 15) = 0 \Rightarrow x - 2 = 0 or x + 15 = 0 \Rightarrow x = 2 or x = - 15 \Rightarrow x = 2 (∵ Width cannot be negative) Thus, the width of the path is 2 m.$

Page No 226:

Question 63:

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Answer:

Let the length of the side of the first and the second square be $x$ and y, respectively.

$According to the question : x^{2} + y^{2} = 640 . . . (i) Also, 4 x - 4 y = 64 \Rightarrow x - y = 16 \Rightarrow x = 16 + y$

Putting the value of $x$ in (i), we get:

$x^{2} + y^{2} = 640 \Rightarrow {(16 + y)}^{2} + y^{2} = 640 \Rightarrow 256 + 32 y + y^{2} + y^{2} = 640 \Rightarrow 2 y^{2} + 32 y - 384 = 0 \Rightarrow y^{2} + 16 y - 192 = 0 \Rightarrow y^{2} + (24 - 8) y - 192 = 0 \Rightarrow y^{2} + 24 y - 8 y - 192 = 0 \Rightarrow y (y + 24) - 8 (y + 24) = 0 \Rightarrow (y + 24) (y - 8) = 0 \Rightarrow y = - 24 or y = 8 ∴ y = 8 (∵ S ide cannot be negative) ∴ x = 16 + y = 16 + 8 = 24 m Thus, the sides of the squares are 8 m and 24 m.$

Page No 226:

Question 64:

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than width of a the rectangle. Their areas being equal, find their dimensions.

Answer:

$Let the breadth of rectangle be x cm . According to the question : Side of the square = (x + 4) cm Length of the rectangle = {3 (x + 4)} cm It is given that the areas of the rectangle and square are same. ∴ 3 (x + 4) \times x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = {(x + 4)}^{2} \Rightarrow 3 x^{2} + 12 x = x^{2} + 8 x + 16 \Rightarrow 2 x^{2} + 4 x - 16 = 0 \Rightarrow x^{2} + 2 x - 8 = 0 \Rightarrow x^{2} + (4 - 2) x - 8 = 0 \Rightarrow x^{2} + 4 x - 2 x - 8 = 0 \Rightarrow x (x + 4) - 2 (x + 4) = 0 \Rightarrow (x + 4) (x - 2) = 0 \Rightarrow x = - 4 or x = 2 ∴ x = 2 (∵ The value of x cannot be negative) Thus, t he breadth of the rectangle is 2 cm and length is {3 (2 + 4) =18} cm. Also, the side of the square is 6 cm.$

Page No 226:

Question 65:

A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

Answer:

$Let the length and breadth of the rectangular garden be x and y metre, respectively. Given: x y = 180 sq m ... (i) and 2 y + x = 39 \Rightarrow x = 39 - 2 y Putting the value of x in (i), we get: (39 - 2 y) y = 180 \Rightarrow 39 y - 2 y^{2} = 180 \Rightarrow 39 y - 2 y^{2} - 180 = 0 \Rightarrow 2 y^{2} - 39 y + 180 = 0 \Rightarrow 2 y^{2} - (24 + 15) y + 180 = 0 \Rightarrow 2 y^{2} - 24 y - 15 y + 180 = 0 \Rightarrow 2 y (y - 12) - 15 (y - 12) = 0 \Rightarrow (y - 12) (2 y - 15) = 0 \Rightarrow y = 12 or y = \frac{15}{2} = 7.5 If y = 12, x = 39 - 24 = 15 If y = 7.5, x = 39 - 15 = 24 Thus, the l ength and breadth of the garden are (15 m and 12 m) or (24 m and 7 .5 m), respectively.$

Page No 226:

Question 66:

The area of a right-triangle is 600 cm². If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (x + 10) cm.
$Area of triangle = \frac{1}{2} x (x + 10) = 600 \Rightarrow x (x + 10) = 1200 \Rightarrow x^{2} + 10 x - 1200 = 0 \Rightarrow x^{2} + (40 - 30) x - 1200 = 0 \Rightarrow x^{2} + 40 x - 30 x - 1200 = 0 \Rightarrow x (x + 40) - 30 (x + 40) = 0 \Rightarrow (x + 40) (x - 30) = 0 \Rightarrow x = - 40 or x = 30 \Rightarrow x = 30 [∵ Altitude cannot be negative] Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively. ∵ {(Hypotenuse)}^{2} = {(Altitude)}^{2} + {(Base)}^{2} \Rightarrow {(Hypotenuse)}^{2} = {(30)}^{2} + {(40)}^{2} \Rightarrow {(Hypotenuse)}^{2} = 900 + 1600 = 2500 \Rightarrow {(Hypotenuse)}^{2} = {(50)}^{2} \Rightarrow (Hypotenuse) = 50 Thus, the dimensions of the triangle are :, Hypotenuse = 50 cm Altitude = 30 cm Base = 40 cm$

Page No 226:

Question 67:

The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.

Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times 3 x \times x = 96 (∵ Area = 96 sq m) \Rightarrow \frac{x^{2}}{2} = 32 \Rightarrow x^{2} = 64 \Rightarrow x = \pm 8$

The value of x cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 $\times$ 8 = 24 m), respectively.

Page No 226:

Question 68:

The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.

Answer:

Let the base be $x$ m.
Therefore, the altitude will be $(x + 7)$ m.

Area of a triangle = $\frac{1}{2} \times Base \times Altitude$

$∴ \frac{1}{2} \times x \times (x + 7) = 165 \Rightarrow x^{2} + 7 x = 330 \Rightarrow x^{2} + 7 x - 330 = 0 \Rightarrow x^{2} + (22 - 15) x - 330 = 0 \Rightarrow x^{2} + 22 x - 15 x - 330 = 0 \Rightarrow x (x + 22) - 15 (x + 22) = 0 \Rightarrow (x + 22) (x - 15) = 0 \Rightarrow x = - 22 or x = 15$

The value of $x$ cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

Page No 226:

Question 69:

The hypotenuse of a right-angled triangle is 20 metres. If the difference between the length of the other sides be 4 metres, find the other sides.

Answer:

Let one side of the right-angled triangle be $x$ m and the other side be $(x + 4)$ m.
On applying Pythagoras theorem, we have:

$20^{2} = {(x + 4)}^{2} + x^{2} \Rightarrow 400 = x^{2} + 8 x + 16 + x^{2} \Rightarrow 2 x^{2} + 8 x - 384 = 0 \Rightarrow x^{2} + 4 x - 192 = 0 \Rightarrow x^{2} + (16 - 12) x - 192 = 0 \Rightarrow x^{2} + 16 x - 12 x - 192 = 0 \Rightarrow x (x + 16) - 12 (x + 16) = 0 \Rightarrow (x + 16) (x - 12) = 0 \Rightarrow x = - 16 or x = 12$

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

Page No 226:

Question 70:

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

Answer:

Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively.
Therefore, the hypotenuse will be $(x + 2)$ cm.
$∴ {(x + 2)}^{2} = y^{2} + x^{2} . . . (i) Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. ∴ h = (2 y + 1) \Rightarrow x + 2 = 2 y + 1 \Rightarrow x = 2 y - 1 P utting the value of x in (i), we get : {(2 y - 1 + 2)}^{2} = y^{2} + {(2 y - 1)}^{2} \Rightarrow {(2 y + 1)}^{2} = y^{2} + 4 y^{2} - 4 y + 1 \Rightarrow 4 y^{2} + 4 y + 1 = 5 y^{2} - 4 y + 1 \Rightarrow - y^{2} + 8 y = 0 \Rightarrow y^{2} - 8 y = 0 \Rightarrow y (y - 8) = 0 \Rightarrow y = 8 cm ∴ x = 16 - 1 = 15 cm ∴ h = 16 + 1 = 17 cm$

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

Page No 226:

Question 71:

The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle.

Answer:

Let the shortest side be $x$ m.
Therefore, according to the question:
Hypotenuse = $(2 x - 1)$ m
Third side = $(x + 1)$ m
On applying Pythagoras theorem, we get:

${(2 x - 1)}^{2} = {(x + 1)}^{2} + x^{2} \Rightarrow 4 x^{2} - 4 x + 1 = x^{2} + 2 x + 1 + x^{2} \Rightarrow 2 x^{2} - 6 x = 0 \Rightarrow 2 x (x - 3) = 0 \Rightarrow x = 0 or x = 3$

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = $(2 \times 3 - 1) =$ 5 m
Third side = (3 + 1) = 4 m

Page No 234:

Question 1:

Which of the following is a quadratic equation?
(a) $x^{2} - 3 \sqrt{x} + 2 = 0$
(b) $x + \frac{1}{x} = x^{2}$
(c) $x^{2} + \frac{1}{x^{2}} = 5$
(d) $2 x^{2} - 5 x = (x - 1)^{2}$

Answer:

(d) $2 x^{2} - 5 x = (x - 1)^{2}$

$A quadratic equation is the equation with degree 2 . ∵ 2 x^{2} - 5 x = {(x - 1)}^{2} \Rightarrow 2 x^{2} - 5 x = x^{2} - 2 x + 1 \Rightarrow 2 x^{2} - 5 x - x^{2} + 2 x - 1 = 0 \Rightarrow x^{2} - 3 x - 1 = 0, which is a quadratic equation$

Page No 234:

Question 2:

Which of the following is a quadratic equation?
(a) $(x^{2} + 1) = (2 - x)^{2} + 3$
(b) $x^{3} - x^{2} = (x - 1)^{3}$
(c) $2 x^{2} + 3 = (5 + x) (2 x - 3)$
(d) None of these

Answer:

(b) $x^{3} - x^{2} = (x - 1)^{3}$

$∵ x^{3} - x^{2} = {(x - 1)}^{3} ⇒ x^{3} - x^{2} = x^{3} - 3 x^{2} + 3 x - 1 ⇒ 2 x^{2} - 3 x + 1 = 0, which is a quadratic equation$

Page No 234:

Question 3:

Which of the following is not a quadratic equation?
(a) $3 x - x^{2} = x^{2} + 5$
(b) $(x + 2)^{2} = 2 (x^{2} - 5)$
(c) $(\sqrt{2} x + 3)^{2} = 2 x^{2} + 6$
(d) $(x - 1)^{2} = 3 x^{2} + x - 2$

Answer:

(c) $(\sqrt{2} x + 3)^{2} = 2 x^{2} + 6$

$∵ {(\sqrt{2} x + 3)}^{2} = 2 x^{2} + 6 ⇒ 2 x^{2} + 9 + 6 \sqrt{2} x = 2 x^{2} + 6 ⇒ 6 \sqrt{2} x + 3 = 0, which is not a quadratic equation$

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Question 4:

If x = 3 is a solution of the equation $3 x^{2} + (k - 1) x + 9 = 0,$ then k = ?
(a) 11
(b) −11
(c) 13
(d) −13

Answer:

(b) −11
$It is given that x = 3 is a solution of 3 x^{2} + (k - 1) x + 9 = 0; therefore, we have : 3 {(3)}^{2} + (k - 1) \times 3 + 9 = 0 \Rightarrow 27 + 3 (k - 1) + 9 = 0 \Rightarrow 3 (k - 1) = - 36 \Rightarrow (k - 1) = - 12 \Rightarrow k = - 11$

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Question 5:

If one root of the equation $2 x^{2} + a x + 6 = 0$ is 2, then a = ?
(a) 7
(b) −7
(c) $\frac{7}{2}$
(d) $\frac{- 7}{2}$

Answer:

(b) −7
$It is given that one root of the equation 2 x^{2} + a x + 6 = 0 is 2. ∴ 2 \times 2^{2} + a \times 2 + 6 = 0 \Rightarrow 2 a + 14 = 0 \Rightarrow a = - 7$

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Question 6:

The sum of the roots of the equation $x^{2} - 6 x + 2 = 0$ is
(a) 2
(b) −2
(c) 6
(d) −6

Answer:

(c) 6
$Sum of the roots of the equation x^{2} - 6 x + 2 = 0 is α + β = \frac{- b}{a} = \frac{- (- 6)}{1} = 6 , where α and β are the roots of the equation.$

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Question 7:

If the product of the roots of the equation $x^{2} - 3 x + k = 10$ is −2, then the value of k is
(a) −2
(b) −8
(c) 8
(d) 12

Answer:

(c) 8
$It is given that the product of the roots of the equation x^{2} - 3 x + k = 10 is - 2 . The equation can be rewritten as : x^{2} - 3 x + (k - 10) = 0 Product of the roots of a quadratic equation = \frac{c}{a} \Rightarrow \frac{c}{a} = - 2 \Rightarrow \frac{(k - 10)}{1} = - 2 \Rightarrow k = 8$

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Question 8:

The ratio of the sum and product of the roots of the equation $7 x^{2} - 12 x + 18 = 0$
(a) 7 : 12
(b) 7 : 18
(c) 2 : 3
(d) 3 : 2

Answer:

(c) 2 : 3

$Given: 7 x^{2} - 12 x + 18 = 0 ∴ α + β = \frac{12}{7} and α β = \frac{18}{7}, where α and β are the roots of the equation ∴ Ratio of the sum and product of the roots = \frac{12}{7} : \frac{18}{7} = 12 : 18 = 2 : 3$

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Question 9:

If one root of the equation $3 x^{2} - 10 x + 3 = 0 is \frac{1}{3},$ then the other root is
(a) $\frac{- 1}{3}$
(b) $\frac{1}{3}$
(c) −3
(d) 3

Answer:

(d) 3
$Given: 3 x^{2} - 10 x + 3 = 0 One root of the equation is \frac{1}{3} . Let the other root be α . We know that : Product of the roots = \frac{c}{a} \Rightarrow \frac{1}{3} \times α = \frac{3}{3} \Rightarrow α = 3$

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Question 10:

If one root of $5 x^{2} + 13 x + k = 0$ be the reciprocal of the other root, then the value of k is
(a) 0
(b) 1
(c) 2
(d) 5

Answer:

(d) 5
$Let the roots of the equation (5 x^{2} + 13 x + k = 0) be α and \frac{1}{α} . ∴ Product of the roots = \frac{c}{a} \Rightarrow α \times \frac{1}{α} = \frac{k}{5} \Rightarrow 1 = \frac{k}{5} \Rightarrow k = 5$

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Question 11:

If the sum of the roots of the equation $k x^{2} + 2 x + 3 x = 0$ is equal to their product, then the value of k is
(a) $\frac{1}{3}$
(b) $\frac{- 1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{- 2}{3}$

Answer:

(d) $\frac{- 2}{3}$
$Given: k x^{2} + 2 x + 3 k = 0 Sum of the roots = Product of the roots ⇒ \frac{- 2}{k} = \frac{3 k}{k} ⇒ 3 k = - 2 ⇒ k = \frac{- 2}{3}$

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Question 12:

The root of a quadratic equation are 5 and −2. Then, the equation is
(a) $x^{2} - 3 x + 10 = 0$
(b) $x^{2} - 3 x - 10 = 0$
(c) $x^{2} + 3 x - 10 = 0$
(d) $x^{2} + 3 x + 10 = 0$

Answer:

(b) $x^{2} - 3 x - 10 = 0$

$It is given that the roots of the quadratic equation are 5 and - 2 . Then, the equation is: x^{2} - (5 - 2) x + 5 \times (- 2) = 0 ⇒ x^{2} - 3 x - 10 = 0$

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Question 13:

If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is
(a) $x^{2} - 6 x + 6 = 0$
(b) $x^{2} + 6 x - 6 = 0$
(c) $x^{2} - 6 x - 6 = 0$
(d) $x^{2} + 6 x + 6 = 0$

Answer:

(a) $x^{2} - 6 x + 6 = 0$

$Given : Sum of roots = 6 Product of roots = 6 Thus, the equation is: x^{2} - 6 x + 6 = 0$

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Question 14:

Objective Questions (MCQ)

If α and β are the roots of the equation $3 x^{2} + 8 x + 2 = 0$ then $(\frac{1}{α} + \frac{1}{β})$ = ?

(a) $\frac{- 3}{8}$ (b) $\frac{2}{3}$ (c) −4 (d) 4

Answer:

It is given that α and β are the roots of the equation $3 x^{2} + 8 x + 2 = 0$ .

$∴ α + β = - \frac{8}{3}$ and $α β = \frac{2}{3}$
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- \frac{8}{3}}{\frac{2}{3}} = - 4$

Hence, the correct answer is option C.

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Question 15:

The roots of the equation $a x^{2} + b x + c = 0$ will be reciprocal of each other if
(a) a = b
(b) b= c
(c) c = a
(d) none of these

Answer:

(c) c = a
$Let the roots of the equation (a x^{2} + b x + c = 0) be α and \frac{1}{α} . ∴ Product of the roots = α \times \frac{1}{α} = 1 \Rightarrow \frac{c}{a} = 1 \Rightarrow c = a$

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Question 16:

If the roots of the equation $a x^{2} + b x + c = 0$ are equal, then then c = ?
(a) $\frac{- b}{2 a}$
(b) $\frac{b}{2 a}$
(c) $\frac{- b^{2}}{4 a}$
(d) $\frac{b^{2}}{4 a}$

Answer:

(d) $\frac{b^{2}}{4 a}$

$It is given that the roots of the equation (a x^{2} + b x + c = 0) are equal. ∴ (b^{2} - 4 a c) = 0 \Rightarrow b^{2} = 4 a c \Rightarrow c = \frac{b^{2}}{4 a}$

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Question 17:

If the equation 9x² + 6kx + 4 = 0 has equal roots then k = ?
(a) 2 or 0
(b) −2 or 0
(c) 2 or −2
(d) 0 only

Answer:

(c) 2 or −2
$It is given that the roots of the equation (9 x^{2} + 6 k x + 4 = 0) are equal. ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(6 k)}^{2} - 4 \times 9 \times 4 = 0 \Rightarrow 36 k^{2} = 144 \Rightarrow k^{2} = 4 \Rightarrow k = \pm 2$

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Question 18:

If the equation $x^{2} + 2 (k + 2) x + 9 k = 0$ has equal rots, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

Answer:

(a) 1 or 4

$It is given that the roots of the equation (x^{2} + 2 (k + 2) x + 9 k = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {2 (k + 2)}^{2} - 4 \times 1 \times 9 k = 0 \Rightarrow 4 (k^{2} + 4 k + 4) - 36 k = 0 \Rightarrow 4 k^{2} + 16 k + 16 - 36 k = 0 \Rightarrow 4 k^{2} - 20 k + 16 = 0 \Rightarrow k^{2} - 5 k + 4 = 0 \Rightarrow k^{2} - 4 k - k + 4 = 0 \Rightarrow k (k - 4) - (k - 4) = 0 \Rightarrow (k - 4) (k - 1) = 0 \Rightarrow k = 4 or k = 1$

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Question 19:

If the equation $4 x^{2} - 3 k x + 1 = 0$ has equal roots, then k = ?
(a) $\pm \frac{2}{3}$
(b) $\pm \frac{1}{3}$
(c) $\pm \frac{3}{4}$
(d) $\pm \frac{4}{3}$

Answer:

(d) $\pm \frac{4}{3}$

$It is given that the roots of the equation (4 x^{2} - 3 k x + 1 = 0) are equal . ∴ (b^{2} - 4 a c) = 0 \Rightarrow {(3 k)}^{2} - 4 \times 4 \times 1 = 0 \Rightarrow 9 k^{2} = 16 \Rightarrow k^{2} = \frac{16}{9} \Rightarrow k = \pm \frac{4}{3}$

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Question 20:

The roots of $a x^{2} + b x + c = 0, a \neq 0$ are real and unequal, if $(b^{2} - 4 a c)$
(a) > 0
(b) = 0
(c) < 0
(d) none of these

Answer:

(a) > 0
$\begin{array}{l} The roots of the equation are real and unequal when (b^{2} - 4 a c) > 0 . \end{array}$

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Question 21:

Objective Questions (MCQ)

In the equation $a x^{2} + b x + c = 0$ , it is given that $D = (b^{2} - 4 a c) > 0$ . Then, the roots of the equation are

(a) real and equal (b) real and unequal
(c) imaginary (d) none of these

Answer:

We know that when discriminant, $D > 0$ , the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

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Question 22:

The roots of the equation $2 x^{2} - 6 x + 7 = 0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(d) imaginary

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 7 = 36 - 56 = - 20 < 0 Thus, the roots of the equation are imaginary .$

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Question 23:

The roots of the equation $2 x^{2} - 6 x + 3 = 0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

Answer:

(b) real, unequal and irrational

$∵ D = (b^{2} - 4 a c) = {(- 6)}^{2} - 4 \times 2 \times 3 = 36 - 24 = 12 12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational .$

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Question 24:

If the roots of $5 x^{2} - k x + 1 = 0$ are real and distinct, then
(a) $- 2 \sqrt{5} < k < 2 \sqrt{5}$
(b) $k > 2 \sqrt{5}$ only
(c) $k < - 2 \sqrt{5}$ only
(d) either $k > 2 \sqrt{5}$ or $k < - 2 \sqrt{5}$

Answer:

(d) either $k > 2 \sqrt{5}$ or $k < - 2 \sqrt{5}$

$It is given that the roots of the equation (5 x^{2} - k x + 1 = 0) are real and distinct . ∴ (b^{2} - 4 a c) > 0 \Rightarrow {(- k)}^{2} - 4 \times 5 \times 1 > 0 \Rightarrow k^{2} - 20 > 0 \Rightarrow k^{2} > 20 \Rightarrow k > \sqrt{20} or k < - \sqrt{20} \Rightarrow k > 2 \sqrt{5} or k < - 2 \sqrt{5}$

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Question 25:

If the equation $x^{2} + 5 k x + 16 = 0$ has no real roots, then
(a) $k > \frac{8}{5}$
(b) $k < \frac{- 8}{5}$
(c) $\frac{- 8}{5} < k > \frac{8}{5}$
(d) none of these

Answer:

(c) $\frac{- 8}{5} < k > \frac{8}{5}$

$It is given that the equation (x^{2} + 5 k x + 16 = 0) has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(5 k)}^{2} - 4 \times 1 \times 16 < 0 \Rightarrow 25 k^{2} - 64 < 0 \Rightarrow k^{2} < \frac{64}{25} \Rightarrow \frac{- 8}{5} < k < \frac{8}{5}$

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Question 26:

If the equation $x^{2} - k x + 1 = 0$ has no real roots, then
(a) k < −2
(b) k > 2
(c) −2 < k < 2
(d) none of these

Answer:

(c) −2 < k < 2

$It is given that the equation x^{2} - k x + 1 = 0 has no real roots. ∴ (b^{2} - 4 a c) < 0 \Rightarrow {(- k)}^{2} - 4 \times 1 \times 1 < 0 \Rightarrow k^{2} < 4 \Rightarrow - 2 < k < 2$

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Question 27:

For what values of k, the equation $k x^{2} - 6 x - 2 = 0$ has real roots?
(a) $k \leq \frac{- 9}{2}$
(b) $k \geq \frac{- 9}{2}$
(c) $k \leq - 2$
(d) None of these

Answer:

(b) $k \geq \frac{- 9}{2}$

$It is given that the roots of the equation (k x^{2} - 6 x - 2 = 0) are real. ∴ D \geq 0 \Rightarrow (b^{2} - 4 a c) \geq 0 \Rightarrow {(- 6)}^{2} - 4 \times k \times (- 2) \geq 0 \Rightarrow 36 + 8 k \geq 0 \Rightarrow k \geq \frac{- 36}{8} \Rightarrow k \geq \frac{- 9}{2}$

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Question 28:

The sum of a number and its reciprocal is $2 \frac{1}{20} .$ The number is
(a) $\frac{5}{4} or \frac{4}{5}$
(b) $\frac{4}{3} or \frac{3}{4}$
(c) $\frac{5}{6} or \frac{6}{5}$
(d) $\frac{1}{6} or 6$

Answer:

(a) $\frac{5}{4} or \frac{4}{5}$

$Let the required number be x . According to the question : x + \frac{1}{x} = \frac{41}{20} \Rightarrow \frac{x^{2} + 1}{x} = \frac{41}{20} \Rightarrow 20 x^{2} - 41 x + 20 = 0 \Rightarrow 20 x^{2} - 25 x - 16 x + 20 = 0 \Rightarrow 5 x (4 x - 5) - 4 (4 x - 5) = 0 \Rightarrow (4 x - 5) (5 x - 4) = 0 \Rightarrow x = \frac{5}{4} or x = \frac{4}{5}$

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Question 29:

The perimeter of a rectangle is 82 m and its area is 400 m². The breadth of the rectangle is
(a) 25 m
(b) 20 m
(c) 16 m
(d) 9 m

Answer:

(c) 16 m

$Let the length and breadth of the rectangle be l and b . Perimeter of the rectangle = 82 m \Rightarrow 2 \times (l + b) = 82 \Rightarrow l + b = 41 \Rightarrow l = (41 - b) . . . (i) Area of the rectangle = 400 m^{2} \Rightarrow l \times b = 400 m^{2} \Rightarrow (41 - b) b = 400 (using (i)) \Rightarrow 41 b - b^{2} = 400 \Rightarrow b^{2} - 41 b + 400 = 0 \Rightarrow b^{2} - 25 b - 16 b + 400 = 0 \Rightarrow b (b - 25) - 16 (b - 25) = 0 \Rightarrow (b - 25) (b - 16) = 0 \Rightarrow b = 25 or b = 16 If b = 25, we have : l = 41 - 25 = 16 Since, l cannot be less than b, ∴ b = 16 m$

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Question 30:

Objective Questions (MCQ)

The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m². The breadth of the field is

(a) 20 m (b) 30 m (c) 12 m (d) 16 m [CBSE 2014]

Answer:

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m²(Given)

$∴ (x + 8) \times x = 240$ (Area = Length × Breadth)
$\Rightarrow x^{2} + 8 x - 240 = 0 \Rightarrow x^{2} + 20 x - 12 x - 240 = 0 \Rightarrow x (x + 20) - 12 (x + 20) = 0 \Rightarrow (x + 20) (x - 12) = 0$
$\Rightarrow x + 20 = 0 or x - 12 = 0 \Rightarrow x = - 20 or x = 12$
∴ x = 12 (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

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Question 31:

Objective Questions (MCQ)

The roots of the quadratic equation $2 x^{2} - x - 6 = 0$ are

(a) $- 2, \frac{3}{2}$ (b) $2, \frac{- 3}{2}$ (c) $- 2, \frac{- 3}{2}$ (d) $2, \frac{3}{2}$ [CBSE 2012]

Answer:

The given quadratic equation is $2 x^{2} - x - 6 = 0$ .

$2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = \frac{- 3}{2}$
Thus, the roots of the given equation are 2 and $\frac{- 3}{2}$ .

Hence, the correct answer is option B.

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Question 32:

The sum of two natural numbers is 8 and their product is 15. Find the numbers. [CBSE 2012]

Answer:

Let the required natural numbers be x and (8 − x).

It is given that the product of the two numbers is 15.
$∴ x (8 - x) = 15 \Rightarrow 8 x - x^{2} = 15 \Rightarrow x^{2} - 8 x + 15 = 0 \Rightarrow x^{2} - 5 x - 3 x + 15 = 0$
$\Rightarrow x (x - 5) - 3 (x - 5) = 0 \Rightarrow (x - 5) (x - 3) = 0 \Rightarrow x - 5 = 0 or x - 3 = 0 \Rightarrow x = 5 or x = 3$

Hence, the required numbers are 3 and 5.

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Question 33:

Very-Short-Answer Questions

Show that x = −3 is a solution of $x^{2} + 6 x + 9 = 0$ . [CBSE 2008]

Answer:

The given equation is $x^{2} + 6 x + 9 = 0$ .

Putting x = −3 in the given equation, we get

LHS = ${(- 3)}^{2} + 6 \times (- 3) + 9 = 9 - 18 + 9 = 0$ = RHS

∴ x = −3 is a solution of the given equation.

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Question 34:

Very-Short-Answer Questions

Show that x = −2 is a solution of $3 x^{2} + 13 x + 14 = 0$ . [CBSE 2008]

Answer:

The given equation is $3 x^{2} + 13 x + 14 = 0$ .

Putting x = −2 in the given equation, we get

LHS = $3 \times {(- 2)}^{2} + 13 \times (- 2) + 14 = 12 - 26 + 14 = 0$ = RHS

∴ x = −2 is a solution of the given equation.

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Question 35:

Very-Short-Answer Questions

If $x = \frac{- 1}{2}$ is a solution of the quadratic equation $3 x^{2} + 2 k x - 3 = 0$ , find the value of k. [CBSE 2015]

Answer:

It is given that $x = \frac{- 1}{2}$ is a solution of the quadratic equation $3 x^{2} + 2 k x - 3 = 0$ .

$∴ 3 \times {(\frac{- 1}{2})}^{2} + 2 k \times (\frac{- 1}{2}) - 3 = 0 \Rightarrow \frac{3}{4} - k - 3 = 0 \Rightarrow k = \frac{3 - 12}{4} = - \frac{9}{4}$

Hence, the value of k is $- \frac{9}{4}$ .

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Question 36:

Very-Short-Answer Questions

Find the roots of the quadratic equation $2 x^{2} - x - 6 = 0$ . [CBSE 2012]

Answer:

The given quadratic equation is $2 x^{2} - x - 6 = 0$ .
$2 x^{2} - x - 6 = 0 \Rightarrow 2 x^{2} - 4 x + 3 x - 6 = 0 \Rightarrow 2 x (x - 2) + 3 (x - 2) = 0 \Rightarrow (x - 2) (2 x + 3) = 0$
$\Rightarrow x - 2 = 0 or 2 x + 3 = 0 \Rightarrow x = 2 or x = - \frac{3}{2}$
Hence, the roots of the given equation are 2 and $- \frac{3}{2}$ .

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Question 37:

Very-Short-Answer Questions

Find the solution of the quadratic equation $3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0$ . [CBSE 2009]

Answer:

The given quadratic equation is $3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0$ .
$3 \sqrt{3} x^{2} + 10 x + \sqrt{3} = 0 \Rightarrow 3 \sqrt{3} x^{2} + 9 x + x + \sqrt{3} = 0 \Rightarrow 3 \sqrt{3} x (x + \sqrt{3}) + 1 (x + \sqrt{3}) = 0 \Rightarrow (x + \sqrt{3}) (3 \sqrt{3} x + 1) = 0$
$\Rightarrow x + \sqrt{3} = 0 or 3 \sqrt{3} x + 1 = 0 \Rightarrow x = - \sqrt{3} or x = - \frac{1}{3 \sqrt{3}} = - \frac{\sqrt{3}}{9}$
Hence, $- \sqrt{3}$ and $- \frac{\sqrt{3}}{9}$ are the solutions of the given equation.

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Question 38:

Very-Short-Answer Questions

If the roots of the quadratic equation $2 x^{2} + 8 x + k = 0$ are equal then find the value of k. [CBSE 2014]

Answer:

It is given that the roots of the quadratic equation $2 x^{2} + 8 x + k = 0$ are equal.
$∴ D = 0 \Rightarrow 8^{2} - 4 \times 2 \times k = 0 \Rightarrow 64 - 8 k = 0 \Rightarrow k = 8$
Hence, the value of k is 8.

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Question 39:

Very-Short-Answer Questions

If the quadratic equation $p x^{2} - 2 \sqrt{5} p x + 15 = 0$ has two equal roots then find the value of p. [CBSE 2015]

Answer:

It is given that the quadratic equation $p x^{2} -$