If the squared difference of the zeros of the quadratic polynomial $1(x)=x^{2}+p x+45$ is equal to 144, find the value of $p$.
Given $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}+p x+45$
$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=\frac{-p}{1}$
= -p
$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{45}{1}$
=45
We have,
$(\alpha-\beta)^{2}=\alpha^{2}+\beta^{2}-2 \alpha \beta$
$144=(\alpha+\beta)^{2}-2 \alpha \beta-2 \alpha \beta$
$144=(\alpha+\beta)^{2}-4 \alpha \beta$
Substituting $\alpha+\beta=-p$ and $\alpha \beta=45$ then we get,
$144=(-p)^{2}-4 \times 4$
$144=p^{2}-4 \times 45$
$144=n^{2}-180$
$144+180=p^{2}$
$324=p^{2}$
$\sqrt{18 \times 18}=p \times p$
$\pm 18=p$
Hence, the value of $p$ is $\pm 18$.