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

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Aicharya · Accepted Answer

(3) $ \begin{aligned} & (1-p)^2+p(1-p)+(1-p)=0 \quad\left(	ext { since }(1-p) 	ext { is a root of the equation } x^2+p x+(1-p)=0ight) \ & \Rightarrow(1-p)(1-p+p+1)=0 \ & \Rightarrow 2(1-p)=0 \Rightarrow(1-p)=0 \Rightarrow p=1 \ & 	ext { sum of root is } \alpha+\beta=-p 	ext { and product } \alpha \beta=1-p=0 \quad 	ext { (where } \beta=1-p=0) \ & \Rightarrow \alpha+0=-1 \Rightarrow \alpha=-1 \Rightarrow 	ext { Roots are } 0,-1 \end{aligned} $

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

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