If (1-p) is a root of x²+px+(1-p)=0, then the roots are (a) -1,2 (b) -1,1 (c) 0,-1 (d) 0,1

Correct answer: (c) 0,-1 Substituting x=1-p: (1-p)(2)=0→ p=1; the equation becomes x²+x=0 with roots 0,-1. JEE Main 2004 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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[JEE Main 2004] If $(1-p)$ is a root of $x^2+px+(1-p)=0$, then the roots are

VAVidaara Admin Asked 1mo ago 2 views 1 answer

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If $(1-p)$ is a root of $x^2+px+(1-p)=0$, then the roots are

(a) $-1,2$
(b) $-1,1$
(c) $0,-1$
(d) $0,1$

1 Answer

VAVidaara Admin ✓ Vidaara Team ✓ Accepted · 1mo ago ▲ 0

Correct answer: (c) $0,-1$

Substituting $x=1-p$: $(1-p)(2)=0\Rightarrow p=1$; the equation becomes $x^2+x=0$ with roots $0,-1$.

JEE Main 2004 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.

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