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

1) 0,1
2) $-1,2$
3) $0,-1$
4) $-1,1$
in Mathematics by Gold Status (10,247 points) | 12 views

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Answer

(3)

Explanations

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

by Gold Status (10,247 points)

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