If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then the value of $\frac{1}{a α + b} + \frac{1}{a β + b}$ is

\frac{a}{b c}

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\frac{b}{a c}

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\frac{c}{a b}

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\frac{a b}{c}

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Solution

The correct option is B $\frac{b}{a c}$
Given:
$a x^{2} + b x + c = 0$
Sum of roots, $S = α + β = \frac{- b}{a}$

Product of roots, $P = α β = \frac{c}{a}$
Now, we have to find the value of $\frac{1}{a α + b} + \frac{1}{a β + b}$
$\frac{1}{a α + b} + \frac{1}{a β + b}$

$= \frac{a β + b + a α + b}{(a α + b) (a β + b)}$

$= \frac{a (β + α) + 2 b}{(a α + b) (a β + b)}$

$= \frac{a (α + β) + 2 b}{a^{2} α β + a b (α + β) + b^{2}}$

Using sum and product of roots here, we get
$\frac{1}{a α + b} + \frac{1}{a β + b}$ $= \frac{- \frac{b}{a} a + 2 b}{a^{2} \times \frac{c}{a} - \frac{b}{a} a b + b^{2}}$
$= \frac{+ b}{a c - b^{2} + b^{2}} = \frac{b}{a c}$

Hence, option B.

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