Question

If $\alpha ,\beta$ are the roots of the equation $x^2-(1+a^2)x+\frac{1}{2}(1+a^2+a^4)=0,$ then ${\alpha }^2+{\beta }^2$ is equal to :

• A. $a^4+a^2$
• B. $a^2$
• C. ${(a^2+a^4)}^2$
• D. None of these

Answer 1

The correct option is A $a^4+a^2$

$x^2-(1+a^2)x+\frac{1}{2}(1+a^2+a^4)=r⟶\left(1\right)$

in a quadratic equation of the form ${ax}^2+bx+c=0$

which has two roots $\alpha$ and $\beta$

$\alpha +\beta =-b/a$

$\alpha \beta =c/a$

Using the above relations we can say that if we have $\alpha ,\beta$ as the roots of equation $\left(1\right)$ then

$\alpha +\beta =(1+a^2)$

$\alpha \beta =+\frac{1}{2}(1+a^2+a^4)$

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

$={(1+a^2)}^2-2\times \frac{1}{2}(1+a^2+a^4)$

$=1+a^4+{2a}^2-1-a^2-a^4$

$=a^2$

$\therefore$ ${\alpha }^2+{\beta }^2=a^2$

Option B.