Question

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

Answer 1

$\alpha +\beta =p,\alpha \beta =q$ ............... (1)

Required to prove: $\frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{p^4}{q^2}-\frac{p^2}{q}+2$

Simplifying the LHS, we get

$\Rightarrow \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{{\left(\alpha \beta \right)}^2}$

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

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

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

From eq (1), we get

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

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

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

Hence proved.