Question

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

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

Answer 1

The correct option is C $0,-1$

Since $(1-p)$ is the root of quadratic equation $x^2+px+(1-p)=0$ ...(1)

So, $(1-p)$ satisfies the above equation

$\therefore {(1-p)}^2+p(1-p)+(1-p)=0$

$\Rightarrow (1-p)(1-p+p+1)=0$

$\Rightarrow (1-p)\left(2\right)=0\Rightarrow p=1$

On putting this value of $p$ in equation (1)
$x^2+x=0\Rightarrow x(x+1)=0\Rightarrow x=0,-1$