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

Answer: 3.41

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