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)

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