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If $\alpha, \beta$ are the roots of the equation $2 x^{2}-2\left(1+n^{2}\right) x+(1+$
$\left.n^{2}+n^{4}\right)=0$, then what is the value of $\alpha^{2}+\beta^{2} ?$
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$\left.n^{2}+n^{4}\right)=0$, then what is the value of $\alpha^{2}+\beta^{2} ?$
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Verified Answer
The correct answer is:
$n^{2}$
Since, $\alpha$ and $\beta$ be the roots of
$2 \mathrm{x}^{2}-2\left(1+n^{2}\right) x+\left(1+n^{2}+n^{4}\right)=0$
$\therefore \quad \alpha+\beta=-\left[\frac{-2\left(1+n^{2}\right)}{2}\right]=\left(n^{2}+1\right)$
and $\alpha \beta=\frac{1+n^{2}+n^{4}}{2}$
Now, Consider $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$=\left(n^{2}+1\right)^{2}-\left(1+n^{2}+n^{4}\right)$
$n^{4}+1+2 n^{2}-1-n^{2}-n^{4}=n^{2}$
$2 \mathrm{x}^{2}-2\left(1+n^{2}\right) x+\left(1+n^{2}+n^{4}\right)=0$
$\therefore \quad \alpha+\beta=-\left[\frac{-2\left(1+n^{2}\right)}{2}\right]=\left(n^{2}+1\right)$
and $\alpha \beta=\frac{1+n^{2}+n^{4}}{2}$
Now, Consider $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$=\left(n^{2}+1\right)^{2}-\left(1+n^{2}+n^{4}\right)$
$n^{4}+1+2 n^{2}-1-n^{2}-n^{4}=n^{2}$
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