Question

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

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

Answer 1

The correct option is A $0,-1$

Substituting $x=1-p$, we get

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

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

Hence, $p=1$

Thus, $x=0$ is root of the equation.

Now, the equation changes to

$x^2+x=0$

$x(x+1)=0$

$x=0$ and $x=-1$