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)

$\left(1-p\right)^{2}+p\left(1-p\right)+\left(1-p\right) = 0\quad$ (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 $α + β = −p$ and product $αβ = 1− p = 0\quad$ (where $β = 1 - p = 0$)
$⇒\quad α + 0 = −1 \quad⇒ α = −1⇒\quad$ Roots are $0, -1$