If one of the root of the equation $a x^{2} + b x + c = 0$ is reciprocal of the one root of the equation $a_{1} x^{2} + b_{1} x + c_{1} = 0$ , then:

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${(a a_{1} - c c_{1})}^{2} = (b c_{1} - b_{1} a) (b_{1} c - a_{1} b)$

None of these

${(b c_{1} - b_{1} c)}^{2} = (c a_{1} - c_{1} a) (a b_{1} - a_{1} b)$

${(a b_{1} - a_{1} b)}^{2} = (b c_{1} - b_{1} c) (c a_{1} - c_{1} a)$

answer is A.

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Detailed Solution

The given quadratic equation is $a x^{2} + b x + c = 0$

If $α$ be the roots of the equation, by the given condition

Then $\frac{1}{α}$ is the roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0$

$∴$ $a α^{2} + b α + c = 0$ …(1)

and $a_{1} \frac{1}{α^{2}} + b_{1} \frac{1}{α} c_{1} = 0$

$\Rightarrow$ $c_{1} α^{2} + b_{1} α + a_{1} = 0$ …(2)

Solving (1) and (2):

$\frac{α^{2}}{b a_{1} - b_{1} c} = \frac{α}{c c_{1} - a a_{1}} = \frac{1}{a b_{1} - b c_{1}}$

$\Rightarrow$ ${(c c_{1} - a a_{1})}^{2} = (b a_{1} - b_{1} c) (a b_{1} - b c_{1})$

Hence ${(a a_{1} - c c_{1})}^{2} = (b c_{1} - b_{1} a) (b_{1} c - a_{1} b)$ .

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