Question

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

Answer 1

Hint: We will be using the concepts of the quadratic equation to solve the problem. We will be using the concepts 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 answer:
Now, we have been given $\alpha $ and $\beta $are the zeroes of a quadratic polynomial $f\left( x \right)={{x}^{2}}-2x+3$and we have to find a polynomial whose roots are $\dfrac{\alpha -1}{\alpha +1},\dfrac{\beta -1}{\beta +1}$.
Now, we know that in a quadratic polynomial $a{{x}^{2}}+bx+c$.
$\begin{align}
  & \text{sum of zeroes}=\dfrac{-b}{a} \\
 & \text{product of zeroes =}\dfrac{c}{a} \\
\end{align}$
Therefore, in polynomial ${{x}^{2}}-2x+3$
$\begin{align}
  & \alpha +\beta =\text{sum of roots}=2............\left( 1 \right) \\
 & \alpha \beta =\text{product of roots }=3............\left( 2 \right) \\
\end{align}$
Now, we know that a polynomial with $\alpha $ and $\beta $as roots can be written as $k\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)$.
Therefore, the polynomial with $\dfrac{\alpha -1}{\alpha +1},\dfrac{\beta -1}{\beta +1}$as roots can be written as;
$\begin{align}
  & k\left( {{x}^{2}}-\left( \dfrac{\alpha -1}{\alpha +1}+\dfrac{\beta -1}{\beta +1} \right)x+\left( \dfrac{\alpha -1}{\alpha +1} \right)\left( \dfrac{\beta -1}{\beta +1} \right) \right) \\
 & =k\left( {{x}^{2}}-\left( \dfrac{\alpha \beta +\alpha -\beta -1+\alpha \beta -\alpha +\beta -1}{\alpha \beta +\alpha +\beta +1} \right)x+\left( \dfrac{\alpha -1}{\alpha +1} \right)\left( \dfrac{\beta -1}{\beta +1} \right) \right) \\
 & =k\left( {{x}^{2}}-\dfrac{\left( 2\alpha \beta -2 \right)}{\left( \alpha \beta +\alpha +\beta +1 \right)}x+\left( \dfrac{\alpha \beta -\left( \alpha +\beta \right)+1}{\alpha \beta +\left( \alpha +\beta \right)+1} \right) \right) \\
\end{align}$
Now, we will substitute the value of $\alpha \beta $ and $\alpha +\beta $from (2) and (1),
Therefore, we have,
$\begin{align}
  & =k\left( {{x}^{2}}-\left( \dfrac{2\left( 3 \right)-2}{3+2+1} \right)x+\left( \dfrac{3-\left( 2 \right)+1}{3+\left( 2 \right)+1} \right) \right) \\
 & =k\left( {{x}^{2}}-\left( \dfrac{6-2}{6} \right)x+\left( \dfrac{2}{6} \right) \right) \\
 & =k\left( {{x}^{2}}-\dfrac{4}{6}x+\dfrac{2}{6} \right) \\
 & =k\left( {{x}^{2}}-\dfrac{2}{3}x+\dfrac{1}{3} \right) \\
 & =k\left( \dfrac{3{{x}^{2}}-2x+1}{3} \right) \\
\end{align}$
Now, since for all values of k the equation will have the same roots. Therefore, we substitute k = 3,
$=3{{x}^{2}}-2x+1$
Therefore, the polynomial is $3{{x}^{2}}-2x+1$.

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