Question

If $\alpha ,\beta ,\gamma ,\delta$ are the roots of $x^4-x^3-7x^2+x+6=0$, then ${\alpha }^4+{\beta }^4+{\gamma }^4+{\delta }^4$ =

• A. $79$
• B. $89$
• C. $99$
• D. $109$

Answer 1

The correct option is C $99$

$x^4-x^3-7x^2+x+6=0$
$x=1$ satisfies the given equation .
$(x-1)(x^3-7x-6)=0$
Again, $x=-1$ satisfies $x^3-7x-6=0$
$\Rightarrow (x-1)(x+1)(x^2-x-6)=0$
$\Rightarrow (x-1)(x+1)(x-3)(x+2)=0$
Hence, the roots are $-1,1,-2,3$
Hence, ${\alpha }^4+{\beta }^4+{\gamma }^4+{\delta }^4=99$