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The value of P for which the sum of squares of roots of equation $x^2-(p-2)x-(p+1)=0$ attains the least value is :

$\begin{array}{1 1}(A)\;0&(B)\;1\\(C)\;2&(D)\;3\end{array} $

Can you answer this question?
 
 

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Let the roots be $\alpha,\beta$
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$
$\Rightarrow (p-2)^2+2(p+1)$
$\Rightarrow p^2-2p+6$
This attains least value at $-\large\frac{b}{2a}$ i.e $p=-\large\frac{(-2)}{2(1)}$$=1$
Hence (B) is the correct answer.
answered Apr 15, 2014 by sreemathi.v
 

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