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If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+c=0$, then what is the value of $(a \alpha+b)^{-1}+(a \beta+b)^{-1} ?$
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The correct answer is:
bac
Since, $\alpha$ and $\beta$ are the roots of the equation $a x^{2}+b x+c=0$, then
Sum of the roots, $\alpha+\beta=-\frac{b}{a}$ and Product of the roots,
$\alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
The expression, $(a \alpha+b)^{-1}+(a \beta+b)^{-1}$
$=\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{\alpha \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$
$=\frac{\mathrm{a}(\alpha+\beta)+2 \mathrm{~b}}{\mathrm{a}^{2} \alpha \beta+\mathrm{ab}(\alpha+\beta)+\mathrm{b}^{2}}$
$=\frac{a(-b / a)+2 b}{a^{2}(c / a)+a b(-b / a)+b^{2}}=\frac{-b+2 b}{a c-b^{2}+b^{2}}=\frac{b}{a c}$
Sum of the roots, $\alpha+\beta=-\frac{b}{a}$ and Product of the roots,
$\alpha \beta=\frac{\mathrm{c}}{\mathrm{a}}$
The expression, $(a \alpha+b)^{-1}+(a \beta+b)^{-1}$
$=\frac{1}{a \alpha+b}+\frac{1}{a \beta+b}=\frac{\alpha \beta+b+a \alpha+b}{(a \alpha+b)(a \beta+b)}$
$=\frac{\mathrm{a}(\alpha+\beta)+2 \mathrm{~b}}{\mathrm{a}^{2} \alpha \beta+\mathrm{ab}(\alpha+\beta)+\mathrm{b}^{2}}$
$=\frac{a(-b / a)+2 b}{a^{2}(c / a)+a b(-b / a)+b^{2}}=\frac{-b+2 b}{a c-b^{2}+b^{2}}=\frac{b}{a c}$
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