$I f o n e r o o t o f t h e e q u a t i o n a x^{2} + b x + c = 0 i s r e c i p r o c a l o f t h e o n e o f t h e r o o t s o f e q u a t i o n a_{1} x^{2} + b_{1} x + c_{1} = 0 t h e n$

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} L e t α b e t h e r o o t s o f a x^{2} + b x + c = 0 \Rightarrow a α^{2} + b α + c = 0 - - (1) \\ a n d \frac{1}{α} b e t h e r o o t o f a_{1} x^{2} + b_{1} x + c_{1} = 0 \\ \Rightarrow a_{1} {(\frac{1}{α})}^{2} + b_{1} (\frac{1}{α}) + c_{1} = 0 \\ \Rightarrow c_{1} α^{2} + b_{1} α + a_{1} = 0 - - (2) \\ (1) \times c_{1} \Rightarrow a c_{1} α^{2} + b c_{1} α + c c_{1} = 0 \\ (2) \times a \Rightarrow a c_{1} α^{2} + a b_{1} α + a a_{1} = 0 \\ \bar{α (b c_{1} - a b_{1}) + c c_{1} - a a_{1} = 0} \\ α = \frac{a a_{1} - c c_{1}}{b c_{1} - a b_{1}} \\ f r o m (1) \\ a {(\frac{a a_{1} - c c_{1}}{b c_{1} - a b_{1}})}^{2} + b (\frac{a a_{1} - c c_{1}}{b c_{1} - a b_{1}}) + c = 0 \\ \Rightarrow a {(a a_{1} - c c_{1})}^{2} + b (a a_{1} - c c_{1}) (b c_{1} - a b_{1}) + c {(b c_{1} - a b_{1})}^{2} = 0 \\ \Rightarrow a {(a a_{1} - c c_{1})}^{2} + (b c_{1} - a b_{1}) (a b a_{1} - b c c_{1} + b c c_{1} - a c b_{1}) = 0 \\ \Rightarrow a {(a a_{1} - c c_{1})}^{2} + (b c_{1} - a b_{1}) a (b a_{1} - c b_{1}) = 0 \\ \Rightarrow {(a a_{1} - c c_{1})}^{2} = (b c_{1} - a b_{1}) (c b_{1} - b a_{1}) \end{array}$