# If α and β are the zeros of the quadratic polynomial f(x)

Question:

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

Solution:

$\frac{\alpha^{2}}{\beta^{2}}+\frac{\beta^{2}}{\alpha^{2}}=\frac{\left[(p)^{2}-2 q\right]^{2}-2(q)^{2}}{q^{2}}$Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $f(x)=x^{2}-p x+q$

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

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

= p

$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{q}{1}$

= q

We have,

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

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

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

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

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

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

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

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

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

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

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

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

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

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