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Relation between Roots and Coefficients for Quadratic

lf the roots ...

Question

lf the roots of the equation $a x^{2} + b x + c = 0$ are the reciprocals of the roots of the equation $p x^{2} + q x + r = 0$ , then

A

a c q^{2} = b^{2} p r

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B

a c = p r

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C

b^{2} a c = q^{2} p r

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D

a b = p q

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Solution

The correct option is A $a c q^{2} = b^{2} p r$
$L e t α, β b e t h e r o o t s o f t h e g i v e n e q u a t i o n a x^{2} + b x + c = 0 ∴ α + β = - \frac{b}{a} . . . . (1) a n d α β = \frac{c}{a} . . . . . (2) N o w t h e e q u a t i o n h a v i n g r o o t s \frac{1}{α}, \frac{1}{β} w i l l b e x^{2} - (\frac{1}{α} + \frac{1}{β}) x + \frac{1}{α β} = 0 \Rightarrow x^{2} - \frac{α + β}{α β} x + \frac{1}{α β} = 0 S u b s t i t u t i n g f o r α + β a n d α β f r o m e q u a t i o n s (1) a n d (2), w e g e t, x^{2} - \frac{- \frac{b}{a}}{\frac{c}{a}} x + \frac{1}{\frac{c}{a}} = 0 \Rightarrow x^{2} + \frac{b}{c} x + \frac{a}{c} = 0 \Rightarrow c x^{2} + b x + a = 0 T h i s e q u a t i o n i s i d e n t i c a l w i t h p x^{2} + q x + r = 0 ∴ c o m p a r i n g t h e c o e f f i c i e n t s o f x^{2}, x a n d c w e h a v e, \frac{c}{p} = \frac{b}{q} = \frac{a}{r} ∴ \frac{c}{p} \times \frac{a}{r} = \frac{b}{q} \times \frac{b}{q} [∵ b o t h \frac{c}{p} a n d \frac{a}{r} a r e e q u a l t o \frac{b}{q}] \Rightarrow a c q^{2} = b^{2} p r A n s - O p t i o n A .$

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Similar questions

Q. The ratio of the roots of the equation

{a x}^{2} + b x + c = 0

is same as the ratio of the roots of equation

{p x}^{2} + q x + r = 0

D_{1}

D_{2}

are the discriminants of

{a x}^{2} + b x + c = 0

{p x}^{2} + q x + r = 0

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α

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\frac{1 - α}{α}

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Q. The ratio of the roots of the equation

a x^{2} + b x + c = 0

is same as the ratio of the roots of the equation

p x^{2} + q x + r = 0

D_{1}

D_{2}

are the discriminants of

a x^{2} + b x + c = 0

p x^{2} + q x + r = 0

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Q. If the roots of the equation

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