If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$, then $\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}$ is equal to
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If $\alpha, \beta$ are the roots of the equation $a x^2+b x+c=0$, then $\frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}$ is equal to

(A) $\frac{2}{a}$

(B) $\frac{2}{b}$

(C) $\frac{2}{c}$

(D) $-\frac{2}{a}$

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SOLUTION —

Here, $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$ and

$\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=\frac{\left(b^2-2 a c\right)}{a^2}$

$\therefore \frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}=\frac{\alpha(a \alpha+b)+\beta(a \beta+b)}{(a \beta+b)(a \alpha+b)}$

$=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}$

$=\frac{a \frac{\left(b^2-2 a c\right)}{a^2}+b\left(-\frac{b}{a}\right)}{a^2\left(\frac{c}{a}\right)+a b\left(-\frac{b}{a}\right)+b^2}=-\frac{2}{a}$

So, The correct option will be (D). 

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