[JEE Advanced 1992] Let $\alpha,\beta$ be the roots of $(x-a)(x-b)=c$, $c\ne0$. Then the roots of $(x-\alpha)(x-\beta)+c=0$ are
Let $\alpha,\beta$ be the roots of $(x-a)(x-b)=c$, $c\ne0$. Then the roots of $(x-\alpha)(x-\beta)+c=0$ are
(a) $a,c$
(b) $b,c$
(c) $a,b$
(d) $a+c,b+c$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (c) $a,b$
$(x-\alpha)(x-\beta)=(x-a)(x-b)-c$, so $(x-\alpha)(x-\beta)+c=(x-a)(x-b)$, whose roots are $a,b$.
JEE Advanced 1992 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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