Question

If the squared difference of the zeros of the quadratic polynomial $f\left(x\right)=x^2+px+45$ is equal to $144$, find the value of $p$.

Answer 1

Let $\alpha ,\beta$ be the zeroes of the given quadratic polynomial $f\left(x\right)=x^2+px+45$

$\therefore \alpha +\beta =-p,\alpha \beta =45$

It is given that, $(\alpha -\beta {)}^2=144$

$\Rightarrow (\alpha +\beta {)}^2–4\alpha \beta =144$

$\Rightarrow (–p{)}^2–4\times 45=144$

$\Rightarrow p^2–180=144$

$\Rightarrow p^2=144+180$

$\Rightarrow p^2=324$

$\Rightarrow p=\pm \sqrt{324}$

$\therefore p=\pm 18$