Question

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

• A. 0, 1
• B. -1, 2
• C. 0, -1
• D. -1, 1

Answer 1

Answer: (C)

0, -1

Answer 2

$\left(1-p\right)^{2}+p\left(1-p\right)+\left(1-p\right) = 0\quad$ (since $\left(1 - p\right)$ is a root of the equation $x^{2} + px + \left(1 - p\right) = 0$) $?\quad \left(1- p\right)\left(1- p + p + 1\right) = 0$ $?\quad 2\left(1- p\right) = 0? \left(1 - p\right) = 0 ? p = 1$ sum of root is $a + ??= -p$ and product $a??= 1- p = 0\quad$ (where $??= 1 - p = 0$) $?\quad a + 0 = -1 \quad? a = -1?\quad$ Roots are $0, -1$