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If the sum of the squares of the roots of $\mathrm{x}^{2}-(\mathrm{p}-2) \mathrm{x}-(\mathrm{p}+1)=0(\mathrm{p} \in \mathrm{R})$ is 5
then what is the value of $\mathrm{p}$ ?
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then what is the value of $\mathrm{p}$ ?
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Verified Answer
The correct answer is:
1
Let $\alpha$ and $\beta$ be the roots of $x^{2}-(p-2) x-(p+1)=0$
Then $\alpha+\beta=(p-2)$ and $\alpha \beta=-(p+1)$
$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$\Rightarrow(\alpha+\beta)^{2}-2 \alpha \beta=5$
$\Rightarrow(p-2)^{2}+2(p+1)=5$
$\Rightarrow \mathrm{p}^{2}-4 \mathrm{p}+4+2 \mathrm{p}+2=5$
$\Rightarrow \mathrm{p}^{2}-2 \mathrm{p}+1=0$
$\Rightarrow(p-1)^{2}=0$
$\Rightarrow \mathrm{p}=1$
Then $\alpha+\beta=(p-2)$ and $\alpha \beta=-(p+1)$
$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$\Rightarrow(\alpha+\beta)^{2}-2 \alpha \beta=5$
$\Rightarrow(p-2)^{2}+2(p+1)=5$
$\Rightarrow \mathrm{p}^{2}-4 \mathrm{p}+4+2 \mathrm{p}+2=5$
$\Rightarrow \mathrm{p}^{2}-2 \mathrm{p}+1=0$
$\Rightarrow(p-1)^{2}=0$
$\Rightarrow \mathrm{p}=1$
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