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

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Question

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

Vedclass · Accepted Answer

$0, -1$

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(C) Given the quadratic equation $x^2 + px + (1 - p) = 0$ $(i)$.Since $(1 - p)$ is a root,it must satisfy the equation:$(1 - p)^2 + p(1 - p) + (1 - p) = 0$Factor out $(1 - p)$:$(1 - p) [(1 - p) + p + 1] = 0$$(1 - p) [2] = 0$This implies $1 - p = 0$,so $p = 1$.Substitute $p = 1$ into the original equation $(i)$:$x^2 + (1)x + (1 - 1) = 0$$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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