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If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ px\ +\ q$, prove that $\frac{α^2}{β^2}\ +\ \frac{β^2}{α^2}\ =\ \frac{p^4}{q^2}\ -\ \frac{4p^2}{q}\ +\ 2$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ px\ +\ q$.
To do:
Here, we have to prove that $\frac{α^2}{β^2}\ +\ \frac{β^2}{α^2}\ =\ \frac{p^4}{q^2}\ -\ \frac{4p^2}{q}\ +\ 2$.
Solution:
We know that,
The standard form of a quadratic equation is $ax^2+bx+c=0$, where a, b and c are
constants and $a≠0$
Comparing the given equation with the standard form of a quadratic equation,
$a=1$, $b=-p$ and $c=q$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-p)}{1}=p$.
Product of the roots $= αβ = \frac{c}{a} = \frac{q}{1}=q$.
LHS
$ \begin{array}{l}
\frac{\alpha ^{2}}{\beta ^{2}} +\frac{\beta ^{2}}{\alpha ^{2}} =\frac{\alpha ^{4} +\beta ^{4}}{\alpha ^{2} \beta ^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( \alpha ^{2}\right)^{2} +\left( \beta ^{2}\right)^{2}}{\alpha ^{2} \beta ^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( \alpha ^{2} +\beta ^{2}\right)^{2} -2\alpha ^{2} \beta ^{2}}{\alpha ^{2} \beta ^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left(( \alpha +\beta )^{2} -2\alpha \beta \right)^{2} -2( \alpha \beta )^{2}}{( \alpha \beta )^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( p^{2} -2( q)\right)^{2} -2( q)^{2}}{( q)^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( p^{2} -2q\right)^{2} -2q^{2}}{q^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{\left( p^{2}\right)^{2} +( 2q)^{2} -2\left( p^{2}\right)( 2q) -2q^{2}}{q^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{p^{4} +4q^{2} -4p^{2} q -2q^{2}}}{q^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{p^{4} +2q^{2} -4p^{2} q}{q^{2}}
\end{array}$
RHS
$ \begin{array}{l}
\frac{p^{4}}{q^{2}} -\frac{4p^{2}}{q} +2=\frac{p^{4} -4p^{2}( q) +2\left( q^{2}\right)}{q^{2}}\\
\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{p^{4} +2q^{2} -4p^{2} q}{q^{2}}
\end{array}$
LHS$=$RHS
Hence proved.
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