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

$0,–1$

To find the roots of equation $x^2+px+(1-p)=0$

Step 1: Calculate the value of $p$

$\begin{array}{ll}⟹(1–p{)}^2+p(1–p)+(1–p)& =0\\ ⟹(1–p)(1–p+p+1)& =0\\ ⟹p& =1\end{array}$

Step 2: Substitute the value of $p$ in given equation.

The given equation reduces to $x^2+x=0$.

$x^2+x=0$

$\Rightarrow x(x+1)=0$

$\text{⇒ either}x=0\text{or}x+1=0$

$⟹x=0\text{or}x=-1$

Hence, the roots of given equation are $0$ and $-1$.Hence, option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is correct option.