The condition that $x^{3} - a x^{2} + b x - c = 0$ may have two of the roots equal to each other but of opposite signs is

a b = c

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\frac{2}{3} a b = c

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a^{2} b = c

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N o n e o f t h e s e

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Solution

The correct option is D $a b = c$
$\begin{matrix} w e h a v e, x^{3} - a x^{2} + b x - c = 0 - - - - - - (i) L e t s, α, - α, β a r e t h e r o o t s o f g i v e n e q u a t i o n : s u m o f r o o t s : \Rightarrow α - α + β = a ∴ β = a N o w, r o o t a i s s a t i f i e d e q u - - - - - (i) \Rightarrow a^{3} - a . a^{2} + b a - c = 0 \Rightarrow a^{3} - a^{3} + b a - c = 0 ∴ a b = c p r o v e s o t h a t t h e c o r r e c t o p t i o n i s A . \end{matrix}$

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If the roots of $a x^{2} + b x + c = 0$ are equal in magnitude but opposite in sign, then a possible condition is

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

x^{3} - p x^{2} + q x - r = 0

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

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