Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2-8x+k$ such that ${\alpha }^2+{\beta }^2=40$, then the value of $k$ is

Answer 1

$p\left(x\right)=x^2-8x+k$

$a=1,b=-8,c=k$

$\alpha +\beta =\frac{-b}{a}=8$, $\alpha \beta =\frac{c}{a}=k$, ${\alpha }^2+{\beta }^2=40$

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

$40=8^2-2k$

$2k=24$

$k=12$