Question

If $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[f\left( x \right) = {{x}^{2}}-px+q\], find the value of \[\dfrac{{{\alpha }^{2}}}{{{\beta }^{2}}}+\dfrac{{{\beta }^{2}}}{{{\alpha }^{2}}}=\dfrac{{{p}^{4}}}{{{q}^{2}}}-\dfrac{4{{p}^{2}}}{q}+2\].

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using sum of roots and product of roots to further simplify the problem.

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $are the zeros of the quadratic polynomial\[f\left( x \right)={{x}^{2}}-px+q\]. We have to find the value of\[\dfrac{{{\alpha }^{2}}}{{{\beta }^{2}}}+\dfrac{{{\beta }^{2}}}{{{\alpha }^{2}}}=\dfrac{{{p}^{4}}}{{{q}^{2}}}-\dfrac{4{{p}^{2}}}{q}+2\].
Now, to find this value we will be using the sum of roots and product of roots. We know that in a quadratic equation $a{{x}^{2}}+bx+c=0$.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$
Therefore, in \[f\left( x \right)={{x}^{2}}-px+q\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=-\left( -p \right)..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=q...........\left( 2 \right) \\
\end{align}$
Now, we will take LHS and prove it to be equal to RHS.
In LHS we have,
\[\dfrac{{{\alpha }^{2}}}{{{\beta }^{2}}}+\dfrac{{{\beta }^{2}}}{{{\alpha }^{2}}}=\dfrac{{{\alpha }^{4}}+{{\beta }^{4}}}{{{\left( \alpha \beta \right)}^{2}}}.........\left( 3 \right)\]
Now, we know that,
${{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}}$
Substituting $a=\alpha \ and\ b=\beta $, we have,
$\begin{align}
  & {{\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)}^{2}}={{\alpha }^{4}}+{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}} \\
 & {{\alpha }^{4}}+{{\beta }^{4}}={{\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)}^{2}}-2{{\alpha }^{2}}{{\beta }^{2}}.........\left( 4 \right) \\
\end{align}$
Also, we know that,
${{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
Therefore,
${{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}}$
Now, substituting this in (4) we have,
${{\alpha }^{4}}+{{\beta }^{4}}={{\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right)}^{2}}-2{{\left( \alpha \beta \right)}^{2}}$
Also, we will substitute this value of ${{\alpha }^{4}}+{{\beta }^{4}}$ in (3). So, that we have,
\[\dfrac{{{\alpha }^{2}}}{{{\beta }^{2}}}+\dfrac{{{\beta }^{2}}}{{{\alpha }^{2}}}=\dfrac{{{\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right)}^{2}}-2{{\left( \alpha \beta \right)}^{2}}}{{{\left( \alpha \beta \right)}^{2}}}\]
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) and (2). So, we have,
$\begin{align}
  & =\dfrac{{{\left( {{\left( p \right)}^{2}}-2q \right)}^{2}}-2{{\left( q \right)}^{2}}}{{{q}^{2}}} \\
 & =\dfrac{{{\left( {{p}^{2}}-2q \right)}^{2}}-2{{q}^{2}}}{{{q}^{2}}} \\
\end{align}$
Now, we will use the algebraic identity that
${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
So, that we have,
$\begin{align}
  & =\dfrac{{{p}^{4}}+4{{q}^{2}}-4{{p}^{2}}q-2{{q}^{2}}}{{{q}^{2}}} \\
 & =\dfrac{{{p}^{4}}-4{{p}^{2}}q+2{{q}^{2}}}{{{q}^{2}}} \\
 & \dfrac{{{\alpha }^{2}}}{{{\beta }^{2}}}+\dfrac{{{\beta }^{2}}}{{{\alpha }^{2}}}=\dfrac{{{p}^{4}}}{{{q}^{2}}}-\dfrac{4{{p}^{2}}}{q}+2 \\
\end{align}$
Since we have LHS = RHS.
Hence Proved.

Note: To solve these types of questions one must know how to find the relation between sum of roots, product of roots and coefficient of quadratic equation.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$