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If $\alpha\:\:and\:\:\beta$ are the roots of $2x^2-2(1+n^2)x+(1+n^2+n^4)=0$, then the value of $\alpha^2+\beta^2=?$

$\begin{array}{1 1} 2 \\ n^2 \\ 2n^2 \\ 2n^4 \end{array}$

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  • If $ax^2+bx+c=0$, then the sum of the roots$=\large\frac{-b}{a}$ and the product of the roots $ =\large\frac{c}{a}$
$2x^2-2(1+n^2)x+(1+n^2+n^4)=0$
$\alpha,\:\beta$ are the roots of the eqn.
$\Rightarrow\:\alpha+\beta=\large\frac{2(1+n^2)}{2}$$=1+n^2$ and $\alpha\beta=\large\frac{1+n^2+n^4}{2}$
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$\Rightarrow\:\alpha^2+\beta^2=(1+n^2)^2-(1+n^2+n^4)$
$=1+n^4-2n^2-1-n^2-n^4=n^2$
answered Jul 12, 2013 by rvidyagovindarajan_1
 

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