Question

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

Answer 1

We have, $f\left(x\right)=x^2-px+q$

Since $\alpha and\beta$ are zeros of the given polynomial, then

$\alpha +\beta =\frac{-coefficientofx}{coefficientof{x}^2}=\frac{-(-p)}{1}=p$ ...(i)

Also, $\alpha \beta =\frac{-constantterm}{coefficientof{x}^2}=\frac{q}{1}=q$ ...(ii)

Now, $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{(\alpha \beta {)}^2}$

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

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

$=\frac{(p^2-2q{)}^2-2q^2}{q^2}$ ($from\left(i\right),\left(ii\right)$)

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

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

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

Hence, proved.