Question

If $\alpha$ and $\beta$ are roots of the quadratic equation $2p^2{x}^2+2p^{3}x-1=0,p\in R-0$, then minimum value of ${\alpha }^4+{\beta }^4$ is

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

Answer 1

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

$2p^2{x}^2+2p^{3}x-1=0$

Here,

$a=2p^2,b=2p^3,c=-1$

$\because \alpha$ and $\beta$ are the roots of given quadratic equation.

$\therefore$ Sum of roots $=\frac{-b}{a}$

$\Rightarrow \alpha +\beta =\frac{-2p^3}{2p^2}=-p$

Product of roots $=\frac{c}{a}$

$\alpha \beta =\frac{-1}{2p^2}$

Using identity: ${(a+b)}^2=a^2+b^2+2ab$, we have

${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

${(-p)}^2={\alpha }^2+{\beta }^2+2\left(\frac{-1}{2p^2}\right)$

$\Rightarrow {\alpha }^2+{\beta }^2=p^2+\frac{1}{p^2}$

Again using identity: ${(a+b)}^2=a^2+b^2+2ab$, we have

${({\alpha }^2+{\beta }^2)}^2={\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2$

$\Rightarrow {(p^2+\frac{1}{p^2})}^2={\alpha }^4+{\beta }^4+2{\left(\frac{-1}{2p^2}\right)}^2$

$\Rightarrow {\alpha }^4+{\beta }^4=p^4+\frac{1}{p^4}+2-\frac{1}{2p^4}$

$\Rightarrow {\alpha }^4+{\beta }^4=p^4+\frac{1}{2p^4}+2$

$\Rightarrow {\alpha }^4+{\beta }^4={(p^2-\frac{1}{\sqrt{2}{p}^2})}^2+\sqrt{2}+2$

For minimum value, ${(p^2-\frac{1}{\sqrt{2}{p}^2})}^2=0$

$\therefore {\alpha }^4+{\beta }^4=2+\sqrt{2}$

Hence the minimum value of ${\alpha }^4+{\beta }^4$ will be $(2+\sqrt{2})$.

Hence the required answer is $\left(B\right)2+\sqrt{2}$.