Question

If $\alpha$ and $\beta$ be the roots of the equation $x^2+px-\frac{1}{2p^2}=0,$ where $p\in R,$ then the minimum value of ${\alpha }^4+{\beta }^4$ is

• A. $\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $2-\sqrt{2}$
• D. $-\sqrt{2}$

Answer 1

The correct option is B $2+\sqrt{2}$

Let $z={\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2$
$\Rightarrow z=\left[\right(\alpha +\beta {)}^2-2\alpha \beta {]}^2-2(\alpha \beta {)}^2$
$={\left[\right(-p{)}^2-2(-\frac{1}{2p^2})]}^2-2\left(\frac{1}{4p^4}\right)$
$={[p^2+\frac{1}{p^2}]}^2-\left(\frac{1}{2p^4}\right)$
$=p^4+2+\frac{1}{p^4}-\frac{1}{2p^4}=2+p^4+\frac{1}{2p^4}$
$=2+2\times p^2\times \frac{1}{\sqrt{2}{p}^2}+(p^2{)}^2+{\left(\frac{1}{\sqrt{2}{p}^2}\right)}^2-2\times p^2\times \frac{1}{\sqrt{2}{p}^2}$
$=2+\sqrt{2}+{(p^2-\frac{1}{\sqrt{2}{p}^2})}^2\ge 2+\sqrt{2}$
Since square of any real number is always non negative real number.