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Find the value of a such that the sum of the squares of the roots of the equation $\;x^{2}-(a-2)x-(a+1)=0\;$ is least .

$(a)\;\pm3\qquad(b)\;\pm6\qquad(c)\;1\qquad(d)\;4$

Can you answer this question?
 
 

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Answer : 1
Explanation :
Let $\;\alpha \;, \beta\;$ be the roots of the equation
Therefore , $\;\alpha+\beta =a-2\;$ and $\; \alpha\beta=-(a+1)$
$\alpha^{2} + \beta^{2} = (\alpha+\beta)^{2} - 2 \alpha \beta$
$= (a-2)^{2} + 2(a+1)$
$= (a-1)^{2}+5$
Therefore , $\;\alpha^{2} + \beta^{2}\;$ will be minimum if $\;(a-1)^{2} =0$
i.e $\;a=1$
answered Apr 17, 2014 by yamini.v
 

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