Question

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

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

Answer 1

The correct option is C $0,-1$

Since, $(1-p)$ is a root of quadratic equation $x^2+px+(1-p)=0$ ....(i)
So, $(1-p)$ satisfied the above equation
${(1-p)}^2+p(1-p)+(1-p)=0$
$\Rightarrow (1-p)(1-p+p+1)=0$
$\Rightarrow 2(1-p)=0$
$\Rightarrow p=1$
On putting $p=1$ in equation (i), we get
$x^2+x=0$
$\Rightarrow x(x+1)=0$

$\Rightarrow x=0,-1$