Solve the following quadratic equation using quadratic formula . 9 x 2 − 9 ( a + b ) x + ( 2 a 2 + 5 a b + 2 b 2 ) = 0 Options: A . &nbsp The roots are 2 a + b 3 and a + 2 b 3 B . &nbsp The roots are 2 a + b 3 and a − 2 b 3 C . &nbsp The roots are 5 a + b 3 and a + 2 b 3 D . &nbsp The roots are 2 a + b 3 and a − 2 b 4

Answer: Option A : A We have, 9 x 2 − 9 ( a + b ) x + ( 2 a 2 + 5 a b + 2 b 2 ) = 0 Comparing this equation with A x 2 + B x + C = 0 , we have A = 9 , B = − 9 ( a + b ) a n d C = 2 a 2 + 5 a b + 2 b 2 ∴ D = B 2 − 4 A C ⇒ D = 81 ( a + b ) 2 − 36 ( 2 a 2 + 5 a b + 2 b 2 ) ⇒ D = 81 ( a 2 + b 2 + 2 a b ) − ( 72 a 2 + 180 a b + 72 b 2 ) ⇒ D = 9 a 2 + 9 b 2 − 18 a b ⇒ D = 9 ( a 2 + b 2 − 2 a b ) ⇒ D = 9 ( a − b ) 2 ≥ 0 ⇒ D ≥ 0 So, the roots of the given equation are real and are given by α = − B + √ D 2 A = 9 ( a + b ) + 3 ( a − b ) 18 = 12 a + 6 b 18 = 2 a + b 3 and, β = − B − √ D 2 A = 9 ( a + b ) − 3 ( a − b ) 18 = 6 a + 12 b 18 = a + 2 b 3

Solve the following quadratic equation using quadratic formula . 9x2&#x

Question

Solve the following quadratic equation using quadratic formula .
$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

Options:

A . The roots are

\frac{2 a + b}{3}

and

\frac{a + 2 b}{3}

B . The roots are

\frac{2 a + b}{3}

and

\frac{a - 2 b}{3}

C . The roots are

\frac{5 a + b}{3}

and

\frac{a + 2 b}{3}

D . The roots are

\frac{2 a + b}{3}

and

\frac{a - 2 b}{4}

Answer: Option A
:
A

We have,
$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$
Comparing this equation with $A x^{2} + B x + C = 0$ , we have
$A = 9, B = - 9 (a + b) a n d C = 2 a^{2} + 5 a b + 2 b^{2}$
$∴ D = B^{2} - 4 A C$
$\Rightarrow D = 81 (a + b)^{2} - 36 (2 a^{2} + 5 a b + 2 b^{2})$
$\Rightarrow D = 81 (a^{2} + b^{2} + 2 a b) - (72 a^{2} + 180 a b + 72 b^{2})$
$\Rightarrow D = 9 a^{2} + 9 b^{2} - 18 a b$
$\Rightarrow D = 9 (a^{2} + b^{2} - 2 a b)$
$\Rightarrow D = 9 (a - b)^{2} \geq 0$
$\Rightarrow D \geq 0$
So, the roots of the given equation are real and are given by
$α = \frac{- B + \sqrt{D}}{2 A} = \frac{9 (a + b) + 3 (a - b)}{18} = \frac{12 a + 6 b}{18} = \frac{2 a + b}{3}$
and, $β = \frac{- B - \sqrt{D}}{2 A} = \frac{9 (a + b) - 3 (a - b)}{18} = \frac{6 a + 12 b}{18} = \frac{a + 2 b}{3}$