Question

If $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[f\left( x \right)={{x}^{2}}-3x-2\], find a quadratic polynomial whose zeros are $\dfrac{1}{2\alpha +\beta }\ and\ \dfrac{1}{2\beta +\alpha }$.

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using the concept of zeros of polynomials to find the sum of zeroes and product of zeroes also we will be using the method of representing a quadratic polynomial in terms of their roots.

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $ are the zeros of the polynomial\[f\left( x \right)={{x}^{2}}-3x-2\].
Now, we know that in a quadratic polynomial $a{{x}^{2}}+bx+c$.
$\begin{align}
  & \text{sum of zeros }=\dfrac{-b}{a} \\
 & \text{product of zeros }=\dfrac{c}{a} \\
\end{align}$
Therefore, in polynomial\[f\left( x \right)={{x}^{2}}-3x-2\]
$\begin{align}
  & \alpha +\beta =\text{sum of roots }=3..........\left( 1 \right) \\
 & \alpha \beta \ \text{=}\ \text{product of roots }=-2...........\left( 2 \right) \\
\end{align}$
Now, we know that a polynomial with $\alpha \ and\ \beta $ as roots can be represented as $k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)$
Therefore, the polynomial with $\dfrac{1}{2\alpha +\beta }\ and\ \dfrac{1}{2\beta +\alpha }$ as roots is,
$\begin{align}
  & ={{x}^{2}}-\left( \dfrac{1}{2\alpha +\beta }+\ \dfrac{1}{2\beta +\alpha } \right)x+\left( \dfrac{1}{2\alpha +\beta } \right)\left( \dfrac{1}{2\beta +\alpha } \right) \\
 & \\
\end{align}$
Now, on further simplifying we have,
$\begin{align}
  & ={{x}^{2}}-\left( \dfrac{2\beta +\alpha +2\alpha +\beta }{\left( 2\alpha +\beta \right)\left( 2\beta +\alpha \right)} \right)+\left( \dfrac{1}{\left( 2\alpha +\beta \right)\left( 2\beta +\alpha \right)} \right) \\
 & =k\left[ {{x}^{2}}-\left( \dfrac{2\alpha +3\beta }{4\alpha \beta +2{{\alpha }^{2}}+2{{\beta }^{2}}+\alpha \beta } \right)+\dfrac{1}{4\alpha \beta +2{{\alpha }^{2}}+2{{\beta }^{2}}+\alpha \beta } \right] \\
 & =k\left[ {{x}^{2}}-\left( \dfrac{3\left( \alpha +\beta \right)}{2\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)+5\alpha \beta } \right)+\dfrac{1}{4\left( {{\alpha }^{2}}+{{\beta }^{2}} \right)+5\alpha \beta } \right] \\
\end{align}$
Now, we have to find the value of ${{\alpha }^{2}}+{{\beta }^{2}}$. We know that,
${{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta $
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ above to find the value of ${{\alpha }^{2}}+{{\beta }^{2}}$. So,
${{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $
Now, from (1) and (2) we have,
$\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}={{\left( 3 \right)}^{2}}-\left( 2 \right)\left( -2 \right) \\
 & =9+4 \\
 & {{\alpha }^{2}}+{{\beta }^{2}}=13.........\left( 3 \right) \\
\end{align}$
Now, we will substitute the values from (1), (2) and (3). So, that we have the required polynomial as,
$\begin{align}
  & =k\left[ {{x}^{2}}-\left( \dfrac{3\left( 3 \right)}{2\left( 13 \right)+5\left( -2 \right)} \right)x+\dfrac{1}{2\left( 13 \right)+5\left( -2 \right)} \right] \\
 & =k\left[ {{x}^{2}}-\left( \dfrac{9}{26-10} \right)x+\dfrac{1}{26-10} \right] \\
 & =k\left[ {{x}^{2}}-\left( \dfrac{9}{16} \right)x+\dfrac{1}{16} \right] \\
 & =k\left[ \dfrac{1}{16}\left( 16{{x}^{2}}-9x+1 \right) \right] \\
\end{align}$
Since, it is true for all values of k. Therefore, let k = 16 and the required quadratic polynomial is $16{{x}^{2}}-9x+1$.

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