Question

If $\alpha ,\beta$ are the roots of $1+x+x^2=0$ then the value of ${\alpha }^2+{\beta }^2+{\alpha }^{-4}+{\beta }^{-4}=$

Answer 1

$x^2+x+1=0$

$a=b=c=1$

$\alpha +\beta =\frac{-b}{a}=-1$

$\alpha \beta =\frac{c}{a}=1$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =-1$

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

So,

${\alpha }^2+{\beta }^2+{\alpha }^{-4}+{\beta }^{-4}$

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

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

$=-1+\left(\frac{-1}{1}\right)$

$=-2$