Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $p\left(x\right)=x^2-6x+k,$ Find the value of $k$ , such that ${\alpha }^2+{\beta }^2=40$

Answer 1

We have,

$p\left(x\right)=x^2-6x+k$ $...........\left(1\right)$

Since,

$\alpha$ and $\beta$ are the zeroes of the given polynomial.

So,

$\alpha +\beta =6$

$\alpha \beta =k$

Since,

${\alpha }^2+{\beta }^2=40$

We know that

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

$6^2=40+2k$

$36=40+2k$

$-4=2k$

$k=-2$

Hence, this is the answer.