# Solve for x :9x2 - 9( a + b )x + ( 2a2 + 5ab + 2b2 )Solve this by both factorisation and by using quadratic formula.

1.) Using factorization

9x2 -9( a + b ) x + (2a2 + 5ab + 2 b2) = 0

⇒9x2 -9( a + b )x + ( 2a+b ) ( a+2b) =0  (By middle term factorization of 2a2 + 5ab + 2b2=0)

⇒9x2 - ( 6a+3b )x - ( 3a+ 6b ) x + ( 2a+b ) ( a+2b) =0

⇒3x ( 3 x - (2a+b )) - ( a + 2 b ) ( 3x -( 2a+b) ) =0

⇒( 3x -(2a +b) ) ( 3x -(a+2b) ) =0 9x 2 – 9(a + b) x + (2a 2 + 2b 2 + 5ab) = 0

Comparing this equation with the standard equation: Ax 2 + Bx + C = 0

Here, A = 9, B = –9 (a + b), C = (2a 2 +2b 2 + 5ab)

∴ Discriminant, D = B2 – 4AC

⇒ D = {–9 (a + b)}2 – 4 � 9 � (2a 2 + 2b 2 + 5ab)

⇒ D = 81(a + b)2 – 36 (2a 2 + 2b 2 + 5ab)

⇒ D = 81(a 2 + b 2+ 2ab) – 72a 2 – 72b 2 – 180ab

⇒ D = 9a 2 + 9b 2 – 18ab

⇒ D = 9(a 2 – 2ab + b 2)

⇒ D = 9(a – b)2 ≥ 0

As D ≥ 0, therefore, the roots of equation are real. Cheers!!!

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factorization

9x2 -9( a + b ) x + 2a2 + 5ab + 2 b2 = 0

9x2 -9( a + b )x + ( 2a+b ) ( a+2b) =0 By middle term facto of 2a+ 5ab + 2b2.

9x2 - ( 6a+3b )x - ( 3a+ 6b ) x + ( 2a+b ) ( a+2b) =0

3x ( 3 x - 2a-b ) - ( a + 2 b ) ( 3x - 2a-b ) =0

( 3x -2a -b ) ( 3x -a-2b ) =0

x = ( 2a+b)/3 or ( a+2b) /3

(1) 9x 2 – 9(a + bx + 2a 2 + 2b 2 + 5ab = 0

Comparing this equation with the standard equation: Ax 2 + Bx + C = 0

Here, A = 9, B = –9 (a + b), C = 2a 2 +2b 2 + 5ab

∴ Discriminant (D) = B2 – 4AC

⇒ D = {–9 (a + b)}2 – 4 × 9 × (2a 2 + 2b 2 + 5ab)

⇒ D = 81(b)2 – 36 (2a 2 + 2b 2 + 5ab)

⇒ D = 81(a b 2+ 2ab) – 72a 2 – 72b 2 – 180ab

⇒ D = 9a 2 + 9b – 18ab

⇒ D = 9(a 2 – 2ab + b 2)

⇒ D = 9(a – b)2 > 0

So, the roots of equation are real, given by: Hence, roots of the equation are .

(2) Hence, roots of the equation are .

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i didnt ub=nderstand after  2 nd stepppppppp

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