Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $p\left(x\right)=x^2+7x+k$ such that $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{25}{12}$, find the value of k.

Answer 1

We know, $\alpha$ and $\beta$ are the zeros of $p\left(x\right)$, then,

$\alpha +\beta =-7$ and $\alpha \beta =k$

Now, $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{25}{12}$

$\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }=\frac{25}{12}$

$\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }=\frac{25}{12}$

$\frac{(-7{)}^2-2(k)}{k}=\frac{25}{12}$

$12(49-2k)=25k$

$49k=588$

$k=12$

Thus, the value of k is 12.