Question

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

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

Answer 1

The correct option is D $0,1$

given equation is $x^2+px+(1-p)=0$

therefore the product of roots $=\frac{c}{a}=(1-p)/1=(1-p)$

one root is $(1-p)$ another root be x

$\Rightarrow x\times (1-p)=1-p$

$\Rightarrow x=1$

substituting x=1 n the equation we get

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

$\Rightarrow p=1$

so the roots are $1,0$