Question

If $\alpha$ and $\beta$ are the zero of the quadratic polynomial $f\left(x\right)=x^2-px+q$, that $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{p^4}{q^2}-\frac{{4p}^2}{q^2}+2$

Answer 1

We have $\alpha +\beta =\frac{-(-p)}{1}=p$ and $\alpha \beta =\frac{q}{1}=q$

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

$=p^2-2q$ from above

From the question $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}={(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })}^2-2\times \frac{\alpha }{\beta }\times \frac{\beta }{\alpha }$

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

or $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}={\left(\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }\right)}^2-2$

$\Rightarrow \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}={\left(\frac{p^2-2q}{q}\right)}^2-2$

$={(\frac{p^2}{q}-2)}^2-2$

$=\frac{p^4}{q^2}+4-\frac{2p^2}{q}\times 2-2$

$=\frac{p^4}{q^2}-\frac{4p^2}{q}+2$

Hence proved.