Solve each of the following quadratic equations:
$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$

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Solution

(1) Using factorization

$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0 \Rightarrow 12 a b x^{2} - 9 x a^{2} - 8 b^{2} x - 6 a b = 0 \Rightarrow 3 a x (4 b x - 3 a) + 2 b (4 b x - 3 a) = 0 \Rightarrow (3 a x + 2 b) (4 b x - 3 a) = 0 ∴ 3 a x + 2 b = 0 o r 4 b x - 3 a = 0 \Rightarrow x = \frac{- 2 b}{3 a} o r \frac{3 a}{4 b}$

(2) Using quadratic formula

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

Discriminant, $D = b^{2} - 4 a c = [- (9 a^{2} - 8 b^{2})]^{2} - 4 (12 a b) (- 6 a b) = 81 a^{4} + 144 a^{2} b^{2} + 64 b^{4} + 288 a^{2} b^{2} = 81 a^{4} + 144 a^{2} b^{2} + 64 b^{4} = (9 a^{2} + 8 b^{2})^{2} \geq 0$

As $D \geq 0$ , therefore, the roots are real.

$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{(9 a^{2} - 8 b^{2}) \pm \sqrt{(9 a^{2} + 8 b^{2})^{2}}}{2 (12 a b)} = \frac{(9 a^{2} - 8 b^{2}) \pm (9 a^{2} + 8 b^{2})}{24 a b} = \frac{(9 a^{2} - 8 b^{2}) + (9 a^{2} + 8 b^{2})}{24 a b} o r \frac{(9 a^{2} - 8 b^{2}) - (9 a^{2} + 8 b^{2})}{24 a b} = \frac{9 a^{2} - 8 b^{2} + 9 a^{2} + 8 b^{2}}{24 a b} o r \frac{9 a^{2} - 8 b^{2} - 9 a^{2} - 8 b^{2}}{24 a b} = \frac{18 a^{2}}{24 a b} o r \frac{- 16 b^{2}}{24 a b} = \frac{3 a}{4 b} o r \frac{- 2 b}{3 a}$

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Similar questions

Q. Solve for

x

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

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

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0, w h e r e a \neq 0 a n d b \neq 0$

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

, where

a \neq 0 a n d b \neq 0

Q. Evaluate

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

, where

a \neq 0

and

b \neq 0

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Solve each of the following quadratic equations: 12abx2−(9a2−8b2)x−6ab=0

Solve each of the following quadratic equations:
$12 a b x^{2} - (9 a^{2} - 8 b^{2}) x - 6 a b = 0$