Question

If $\alpha$ and $\beta$ the roots of the equation $x^{2} +px-1/ (2p^{2}) = 0,$ where $P\in R$ Then the minimum value of $\alpha^{4} +\beta^{4}$ is______.

Answer 1

Here, $\alpha^{4}+\beta^{4} =\left(\alpha^{2}+\beta^{2}\right)^{2} -2\alpha^{2} \beta^{2}$
$=\left\{\left(\alpha+\beta\right)^2 - 2\alpha \beta\right\}^{2}-2\left(\alpha\beta\right)^{2}$
$=\left(p^{2}+\frac{1}{p^{2}}\right)^{2} -\frac{1}{2p^{4}}$
$=p^{4}+\frac{1}{2p^{4}}+2 $
$=\left(p^{2}=\frac{1}{\sqrt{2}P^{2}}\right)^{2}+2+\sqrt{2}\ge2+\sqrt{2} $
Thus, the minimum value of $\alpha^{4}+\beta^{4}$ is $2+\sqrt{2}$