If one root of the quadratic equation x^2 + p | vedclass

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Question

(A) 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

$0, -1$

Vedclass · Answer

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

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

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