Solve for $x : a x^{2} + (4 a^{2} - 3 b) x - 12 a b = 0$ .

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Solution

Given:
$a x^{2} + (4 a^{2} - 3 b) x - 12 a b = 0$
$\Rightarrow$ $a x^{2} + 4 a x^{2} - 3 b x - 12 a b = 0$
$\Rightarrow$ $a x (x + 4 a) - 3 b (x + 4 a) = 0$
$\Rightarrow$ $(a x - 3 b) (x + 4 a) = 0$
$\Rightarrow$ $a x - 3 b = 0, x + 4 a = 0$
$\Rightarrow$ $a x = 3 b, x = - 4 a$
$\Rightarrow$ $x = \frac{3 b}{a}, x = - 4 a$
Hence, $x = \frac{3 b}{a}$ or $x = - 4 a$ .

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