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

$\alpha$ and $\beta$ the roots of $x^2-px+q$

$\Rightarrow \alpha +\beta =p$, $\alpha +\beta =q$

$\Rightarrow (\alpha +\beta {)}^2=p^2$

${\alpha }^2+{\beta }^2+2\alpha \beta =p^2$

${\alpha }^2+{\beta }^2=p^2-2q$ $\sin \alpha \beta =V$

$({\alpha }^2+{\beta }^2)(p^2-2q{)}^2$

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

${\alpha }^4+{\beta }^4+2q^2=p^4+4q^2-4p^{2}q$......(1)

To prove

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

$LHS$

$\frac{{\alpha }^4+{\beta }^4}{{\alpha }^2{\beta }^2}\Rightarrow \frac{p^4+4q^2-4p^{2}q-2q^2}{q^2}$

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

$\Rightarrow \frac{p^4}{q^2}-\frac{4p^2}{q}+2\to RHS$

Hence $LHS=RHS$