For what value of lambda the sum of the squar | vedclass

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Question

(c) Given equation is ${x^2} + (2 + \lambda )x - \frac{1}{2}(1 + \lambda ) = 0$

So, $\alpha + \beta = - (2 + \lambda ) = 0$ and $\alpha \beta = - \

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(c) Given equation is ${x^2} + (2 + \lambda )x - \frac{1}{2}(1 + \lambda ) = 0$

So, $\alpha + \beta = - (2 + \lambda ) = 0$ and $\alpha \beta = - \left( {\frac{{1 + \lambda }}{2}} \right)$

Now, ${\alpha ^2} + {\beta ^2} = {(\alpha + \beta )^2} - 2\alpha \beta $

$ \Rightarrow {\alpha ^2} + {\beta ^2} = {\left[ { - (2 + \lambda )} \right]^2} + 2.\frac{{(1 + \lambda )}}{2}$

==> ${\alpha ^2} + {\beta ^2} = {\lambda ^2} + 4 + 4\lambda + 1 + \lambda = {\lambda ^2} + 5\lambda + 5$

which is minimum for $\lambda = 1/2$.

For what value of $\lambda$ the sum of the squares of the roots of ${x^2} + (2 + \lambda )\,x - \frac{1}{2}(1 + \lambda ) = 0$ is minimum

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