show that the equation 2(a2 + b2 ) x2 + 2(a+b)x + 1 =0 has no real roots , when a not equal to b .

Please reply soon.

I HAVE ALREADY CHECKED THE EARLIER ANSWERS PROVIDED BY U BUT THERE WAS A MISTAKE IN THAT

 let the roots be alpha and 3alpha

alpha+3alpha = -b/a

4alpha = -b/a

alpha = -b/4a ------(1)

alpha*3alpha = c/a

3(alpha)^2 = c/a -----(2)

substitute (1) in (2)

3(-b/4a)^2 = c/a

b^2/16a^2 = c/3a

b^2 = (16a^2*c)/3a

b^2 = 16ac/3




b^2 : ac = 16ac/3 : ac = 16ac : 3ac = 16 : 3
 

For real roots, D ≥ 0.

For real roots, D ≥ 0.

Equations (1) and (2) are both satisfied simultaneously when

Given, roots of both the equations are real.

First equation:

ax² + 2bx + c = 0

Its discriminant, D ≥ 0

⇒ (2b)² - 4(ac) ≥ 0

⇒ 4b² - 4ac ≥ 0

⇒ 4b² ≥ 4ac

⇒ b² ≥ ac ... (1)

Second equation:

bx² - 2√(ac) x + b = 0

Its discriminant, D ≥ 0

⇒ (2√(ac))² - 4(b²) ≥ 0

⇒ 4ac - 4b² ≥ 0

⇒ 4ac ≥ 4b²

⇒ ac ≥ b² ... (2)

The results of equation (1) and (2) are simultaneously possible in only one case when b² = ac

b ^{2 -} 4ac = 4(a+b)² - 8(a²+b²)

=4a² + 4b² +8ab - 8a² - 8b²

= - 4a² - 4b² +8ab

= -4(a²+b² - 2ab)

= -4(a -b)²

if a is not = 0 then b ^{2 -} 4ac < 0

the roots are not real

2(a ² + b ² ) x ² + 2(a+b)x + 1 =0

On comparing with the general quadratic equation Ax ² + Bx + C = 0

For no real roots, B ² -4AC < 0

⇒[2(a+b)] ² - [4 x 2(a ² +b ² ) x 1] <0

⇒[2a+2b] ² - [8(a ² +b ² )] <0

⇒4a ² +4b ² - 8a ² -8b ² <0

⇒-4a ² -4b ² <0

⇒-4(a ² +b ² ) <0

⇒a ² +b ² < 0

⇒a ² < -b ²

∴a ≠ b.

 plz tell wich 1 should i refer 2

 dude plz tell wich i should refer to

third answer it is right

all are wrong i checked as

[2a+2b]²

is 4a2 + 4b2 + 8ab