Question

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

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

Answer 1

The correct option is C $0,-1$

$x^2+px+(1-p)=0$
Sum of the roots
$=\frac{-b}{a}$
$=-p$
Now let the other root be $\alpha$
Then
$(1-p)+\alpha =-p$
Or
$1-p+\alpha =-p$
$\alpha +1=0$
Or
$\alpha =-1$
Substituting in the equation
$1-p+(1-p)=0$
$2(1-p)=0$
Or
$p=1$.
If $p=1$ then the constant term becomes 0.
Hence the other root will be 0.
Thus the roots are $(-1,0)$.