Question

If $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[f\left( x \right) = {{x}^{2}}-1\], find a quadratic polynomial whose zeros are $\dfrac{2\alpha }{\beta }\ and\ \dfrac{2\beta }{\alpha }$.

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 polynomial\[f\left( x \right)={{x}^{2}}-1\]. We have to find a quadratic polynomial whose zeros are$\dfrac{2\alpha }{\beta }\ and\ \dfrac{2\beta }{\alpha }$.
Now, we know that in a quadratic polynomial $a{{x}^{2}}+bx+c=0$.
$\begin{align}
  & \text{sum of roots }=\dfrac{-b}{a} \\
 & \text{product of roots }=\dfrac{c}{a} \\
\end{align}$
Therefore, in polynomial\[f\left( x \right)={{x}^{2}}-1\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=\dfrac{-\left( 0 \right)}{1}=0..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=\dfrac{-1}{1}...........\left( 2 \right) \\
\end{align}$
Now, we know that a polynomial with $\alpha \ and\ \beta $ as roots can be represented as ${{x}^{2}}-\left( \alpha +\beta \right)x+\left( \alpha +\beta \right)$
Therefore, the polynomial with $\dfrac{2\alpha }{\beta }\ and\ \dfrac{2\beta }{\alpha }$ as roots is,
$\begin{align}
  & ={{x}^{2}}-\left( \dfrac{2\alpha }{\beta }+\dfrac{2\beta }{\alpha } \right)x+\left( \dfrac{2\alpha }{\beta } \right)\left( \dfrac{2\beta }{\alpha } \right) \\
 & ={{x}^{2}}-\left( \dfrac{2{{\alpha }^{2}}}{\beta }+\dfrac{2{{\beta }^{2}}}{\alpha } \right)x+4 \\
 & ={{x}^{2}}-2\dfrac{\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)}{\alpha \beta }x+4 \\
\end{align}$
Now, we know that,
$\begin{align}
  & {{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \\
 & {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta ={{\alpha }^{2}}+{{\beta }^{2}} \\
\end{align}$
Substituting this we have the required polynomial as,
$={{x}^{2}}-2\dfrac{\left( {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \right)}{\alpha \beta }x+4$
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) and (2). So, the polynomial is,
$\begin{align}
  & ={{x}^{2}}-2\dfrac{\left( {{\left( 0 \right)}^{2}}-2\left( -1 \right) \right)}{\left( -1 \right)}x+4 \\
 & ={{x}^{2}}-2\dfrac{\left( 2 \right)}{-1}x+4 \\
 & ={{x}^{2}}+4x+4 \\
\end{align}$
Therefore, the polynomial with $\dfrac{2\alpha }{\beta }\ and\ \dfrac{2\beta }{\alpha }$ as roots is ${{x}^{2}}+4x+4$.

Note: To solve these type of questions one must know that a quadratic polynomial having $\alpha \ and\ \beta $ as roots can be represented as ${{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta $.