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

Solution :
Given equation is
$$\eqalign{ & {x^2} + \left( {2 + \lambda } \right)x - \frac{1}{2}\left( {1 + \lambda } \right) = 0 \cr & {\text{So, }}\alpha + \beta = - \left( {2 + \lambda } \right) = 0{\text{ and }}\alpha \beta = - \left( {\frac{{1 + \lambda }}{2}} \right) \cr & {\text{Now, }}{\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta \cr & \Rightarrow {\alpha ^2} + {\beta ^2} = {\left[ { - \left( {2 + \lambda } \right)} \right]^2} + 2\frac{{\left( {1 + \lambda } \right)}}{2} \cr & \Rightarrow {\alpha ^2} + {\beta ^2} = {\lambda ^2} + 4 + 4\lambda + 1 + \lambda \cr & = {\lambda ^2} + 5\lambda + 5 \cr} $$
Which is minimum for $$\lambda = \frac{1}{2}.$$