Question

step by step don't spamm
if (1-p) is a root of the equestion
{x}^{2} + px + 1 - p = 0x 2+px+1−p=0
then root are​​

Answer 1

Answer:

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Explanation:

We are given that one of the roots of the given equation x2+ px + (1 - p) = 0is (1-p). Let αbe the other root of the equation. Hence, one of the roots of the equation x2+ px + (1 - p) = 0 is -1. Hence, one of the roots of the equation x2+ px + (1 - p) = 0is 0.

Answer 2

$\huge\mathtt\red{Answer}$

We are given that one of the roots of the given equation x2+ px + (1 - p) = 0is (1-p). Let αbe the other root of the equation. Hence, one of the roots of the equation x2+ px + (1 - p) = 0 is -1. Hence, one of the roots of the equation x2+ px + (1 - p) = 0is 0.

I hope it will help you Ankit