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The value of $a$ for which the sum of the squares of the roots of the equation $x^{2}-(a-2) x-a-1=0$ assume the least value is

1) 1
2) 0
3) 3
4) 2
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(1)

Explanation

$x^{2}-(a-2) x-a-1=0$
$\Rightarrow \alpha+\beta=a-2$
$\alpha \beta=-(a+1)$
$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$
$\quad=a^{2}-2 a+6=(a-1)^{2}+5$
$\Rightarrow a=1 .$

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