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Question:

$12 a b x^{2}-\left(9 a^{2}-8 b^{2}\right) x-6 a b=0$, where $a \neq 0$ and $b \neq 0$

 

Solution:

Given:

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

On comparing it with $\mathrm{A} x^{2}+B x+C=0$, we get:

$\mathrm{A}=12 a b, B=-\left(9 a^{2}-8 b^{2}\right)$ and $C=-6 a b$

Discriminant $D$ is given by:

$D=B^{2}-4 A C$

$=\left[-\left(9 a^{2}-8 b^{2}\right)\right]^{2}-4 \times 12 a b \times(-6 a b)$

$=81 a^{4}-144 a^{2} b^{2}+64 b^{4}+288 a^{2} b^{2}$

$=81 a^{4}+144 a^{2} b^{2}+64 b^{4}$

$=\left(9 a^{2}+8 b^{2}\right)^{2}>0$

Hence, the roots of the equation are equal.

Roots $\alpha$ and $\beta$ are given by:

$\alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-\left[-\left(9 a^{2}-8 b^{2}\right)\right]+\sqrt{\left(9 a^{2}+8 b^{2}\right)^{2}}}{2 \times 12 a b}=\frac{9 a^{2}-8 b^{2}+9 a^{2}+8 b^{2}}{24 a b}=\frac{18 a^{2}}{24 a b}=\frac{3 a}{4 b}$

$\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-\left[-\left(9 a^{2}-8 b^{2}\right)\right]-\sqrt{\left(9 a^{2}+8 b^{2}\right)^{2}}}{2 \times 12 a b}=\frac{9 a^{2}-8 b^{2}-9 a^{2}-8 b^{2}}{24 a b}=\frac{-16 b^{2}}{24 a b}=\frac{-2 b}{3 a}$

Thus, the roots of the equation are $\frac{3 a}{4 b}$ and $\frac{-2 b}{3 a}$.

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