If (1 - p) is a root of the quadratic equatio | vedclass

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Question

(B) Given the quadratic equation $x^2 + px + (1 - p) = 0$. Since $(1 - p)$ is a root,it must satisfy the equation: $(1 - p)^2 + p(1 - p) + (1 - p) = 0

Vedclass · Accepted Answer

$0, -1$

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(B) Given the quadratic equation $x^2 + px + (1 - p) = 0$. Since $(1 - p)$ is a root,it must satisfy the equation: $(1 - p)^2 + p(1 - p) + (1 - p) = 0$ $(1 - p) [ (1 - p) + p + 1 ] = 0$ $(1 - p) [ 2 ] = 0$ This implies $1 - p = 0$,so $p = 1$. Substituting $p = 1$ into the original equation: $x^2 + 1x + (1 - 1) = 0$ $x^2 + x = 0$ $x(x + 1) = 0$ Thus,the roots are $x = 0$ and $x = -1$.

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

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