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If $(1-p)$ is a root of quadratic equation $x^2+p x+(1-p)=0$, then its roots are
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Verified Answer
The correct answer is:
$0,-1$
$0,-1$
(3)
$$
\begin{aligned}
& (1-p)^2+p(1-p)+(1-p)=0 \quad\left(\text { since }(1-p) \text { is a root of the equation } x^2+p x+(1-p)=0\right) \\
& \Rightarrow(1-p)(1-p+p+1)=0 \\
& \Rightarrow 2(1-p)=0 \Rightarrow(1-p)=0 \Rightarrow p=1 \\
& \text { sum of root is } \alpha+\beta=-p \text { and product } \alpha \beta=1-p=0 \quad \text { (where } \beta=1-p=0) \\
& \Rightarrow \alpha+0=-1 \Rightarrow \alpha=-1 \Rightarrow \text { Roots are } 0,-1
\end{aligned}
$$
$$
\begin{aligned}
& (1-p)^2+p(1-p)+(1-p)=0 \quad\left(\text { since }(1-p) \text { is a root of the equation } x^2+p x+(1-p)=0\right) \\
& \Rightarrow(1-p)(1-p+p+1)=0 \\
& \Rightarrow 2(1-p)=0 \Rightarrow(1-p)=0 \Rightarrow p=1 \\
& \text { sum of root is } \alpha+\beta=-p \text { and product } \alpha \beta=1-p=0 \quad \text { (where } \beta=1-p=0) \\
& \Rightarrow \alpha+0=-1 \Rightarrow \alpha=-1 \Rightarrow \text { Roots are } 0,-1
\end{aligned}
$$
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