Question

If $\alpha ,\beta$ be the roots of the equation
$2x^2-2(m^2+1)x+(m^4+m^2+1)=0$
then $({\alpha }^2+{\beta }^2)=$

• A. $m$
• B. $0$
• C. $1$
• D. $m^2$

Answer 1

The correct option is D

$m^2$

Given:$2x^2-2(m^2+1)x+(m^4+m^2+1)=0$

$\alpha +\beta =\frac{2(m^2+1)}{2}=m^2+1$ ......(i)

and $\alpha .\beta =\frac{m^4+m^2+1}{2}$ .....(ii)

Using ${\alpha }^2+{\beta }^2$ = $(\alpha +\beta {)}^2-2\alpha \beta$
From $\left(i\right)\text{and}\left(ii\right),$
$\Rightarrow {\alpha }^2+{\beta }^2=(m^2+1{)}^2-2\frac{(m^4+m^2+1)}{2}$
$\Rightarrow {\alpha }^2+{\beta }^2=m^4+2m^2+1-m^4-m^2-1=m^2$