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If $α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$, find a polynomial whose roots are $\frac{α\ -\ 1}{α\ +\ 1},\ \frac{β\ -\ 1}{β\ +\ 1}$.
Given:
$α$ and $β$ are the zeros of the quadratic polynomial $f(x)\ =\ x^2\ -\ 2x\ +\ 3$.
To do:
We have to find the quadratic polynomial having $\frac{α\ -\ 1}{α\ +\ 1},\ \frac{β\ -\ 1}{β\ +\ 1}$ as its zeros.
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, $a=1$, $b=-2$ and $c=3$
Sum of the roots $= α+β = \frac{-b}{a} = \frac{-(-2)}{1}=2$.
Product of the roots $= αβ = \frac{c}{a} = \frac{3}{1}=3$.
Let the sum and product of the zeros of the given quadratic equation be $S$ and $P$ respectively.
Therefore,
$ \begin{array}{l}
S=\frac{\alpha -1}{\alpha +1} +\frac{\beta -1}{\beta +1}\\
\\
\ \ \ =\frac{( \alpha -1)( \beta +1) +( \beta -1)( \alpha +1)}{( \alpha +1)( \beta +1)}\\
\\
\ \ \ =\frac{\alpha \beta +\alpha -\beta -1+\alpha \beta +\beta -\alpha -1}{\alpha \beta +\alpha +\beta +1}\\
\\
\ \ \ =\frac{2\alpha \beta -2}{( \alpha \beta ) +( \alpha +\beta ) +1}\\
\\
\ \ \ =\frac{2( 3) -2}{3+( 2) +1}\\
\\
\ \ \ =\frac{6-2}{6}\\
\\
\ \ \ =\frac{4}{6}\\
\\
\ \ \ =\frac{2}{3}\\
\\
P=\frac{\alpha -1}{\alpha +1} \times \frac{\beta -1}{\beta +1}\\
\\
\ \ \ =\frac{( \alpha -1)( \beta -1)}{( \alpha +1)( \beta +1)}\\
\\
\ \ \ =\frac{\alpha \beta -\alpha -\beta +1}{\alpha \beta +\alpha +\beta +1}\\
\\
\ \ \ =\frac{\alpha \beta -( \alpha +\beta ) +1}{\alpha \beta +( \alpha +\beta ) +1}\\
\\
\ \ \ =\frac{3-( 2) +1}{3+( 2) +1}\\
\\
\ \ \ =\frac{4-2}{6}\\
\\
\ \ =\frac{2}{6}\\
\\
\ \ =\frac{1}{3}
\end{array}$
The quadratic polynomial having the sum of the roots $S$ and product of the roots $P$ is $f(x)=k(x^2-(S)x+P)$, where $k$ is any non-zero real number.
Therefore,
$f(x)=k(x^2-(\frac{2}{3})x+\frac{1}{3})$
$f(x)=k(x^2-\frac{2}{3}x+\frac{1}{3})$
The required quadratic polynomial is $f(x)=k(x^2-\frac{2}{3}x+\frac{1}{3})$, where $k$ is any non-zero real number.
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