Question

If α and β are the zeros of the quadratic polynomial f(x) = x² − px + q, prove that ${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{p^4}{q^2}-\frac{4p^2}{q}+2}$

Answer 1

Since α and β are the zeros of the quadratic polynomial f(x) = x² − px + q

${\displaystyle \alpha +\beta =-\frac{\text{coefficient of x}}{\text{coefficient of}{x}^2}}$

${\displaystyle =\frac{-(-p)}{1}}$

= p

${\displaystyle \alpha \beta =\frac{\text{constant term}}{\text{coefficient of}}{x}^2}$

${\displaystyle =\frac{q}{1}}$

= q

we have,

${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^2\times {\alpha }^2}{{\beta }^2\times {\alpha }^2}+\frac{{\beta }^2\times {\beta }^2}{{\alpha }^2\times {\beta }^2}}$

${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4}{{\beta }^2{\alpha }^2}+\frac{{\beta }^4}{{\alpha }^2{\beta }^2}}$

${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{\alpha }^4+{\beta }^4}{{\alpha }^2{\beta }^2}}$

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

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

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

${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{{[p^2-2q]}^2-2q^2}{q^2}}$

${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}=\frac{[p^2\times p^2-2\times p^2\times 2q+2q\times 2q]-2q^2}{q^2}}$

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

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

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

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

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

Hence, it is proved that ${\displaystyle \frac{{\alpha }^2}{{\beta }^2}+\frac{{\beta }^2}{{\alpha }^2}\text{is equal to}\frac{p^4}{q^2}-\frac{4p^2}{q}+2}$