Solve the given quadratic equation by factorization method
$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

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Solution

We have
$9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

Here Constant term $= 2 a^{2} + 5 a b + 2 b^{2} = 2 a^{2} + 4 a b + a b + 2 b^{2} = 2 a (a + 2 b) + b (a + 2 b) = (2 a + b) (a + 2 b)$

and coefficient of middle term $= - 9 (a + b) = - 3 [(a + b) + (a + 2 b)]$

$∴ 9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0$

$\Rightarrow 9 x^{2} - 3 {(2 a + b) + (a + 2 b)} x + (2 a + b) (a + 2 b) = 0 \Rightarrow 3 x {3 x - (2 a + b)} - (a + 2 b) {3 x - (2 a + b)} = 0$

$\Rightarrow {3 x - (2 a + b)} {3 x - (a + 2 b)} = 0 \Rightarrow {3 x - (2 a + b)} = 0 o r {3 x - (a + 2 b)} = 0$

$\Rightarrow x = \frac{2 a + b}{3} o r x = \frac{a + 2 b}{3}$

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Similar questions

Solve the following quadratic equation using quadratic formula .

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

9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0

9 x^{2} - 9 (a + b) + (2 a^{2} + 5 a b + 2 b^{2}) = 0

x

9 x^{2} - 9 (a + b) x + (2 a^{2} + 5 a b + 2 b^{2}) = 0

9 x^{2} - 9 (a + b) x + (2 a 62 + 5 a b + 2 b^{2}) = 0

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