If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+3=0$. Then find the equation whose roots are $\frac{\alpha -1}{\alpha +1},\frac{\beta -1}{\beta +1}$.

Given equation: $x^2-2x+3=0$

The roots of the equation are $\alpha$ and $\beta$.

We know that, 

The sum of the roots, $\alpha +\beta =\frac{-\mathrm{coefficient} \mathrm{of} x}{\mathrm{coefficient }\mathrm{of }{\mathrm{x}}^2}\mathrm{ }$

$\alpha +\beta =-\frac{(-2)}{1}$

$\alpha +\beta =2$

The product of the roots, $\mathit{\alpha \beta }=\frac{\mathrm{constant} \mathrm{term}}{\mathrm{coefficient }\mathrm{of }{\mathrm{x}}^2}$

$\alpha \beta =\frac{3}{1}$

$\mathit{\alpha \beta }=3$ 

Now, the roots of another equation is $\frac{\alpha -1}{\alpha +1},\frac{\beta -1}{\beta +1}$

Here, the sum of roots, $\frac{\alpha -1}{\alpha +1}+\frac{\beta -1}{\beta +1}$

$=\frac{(\alpha -1)(\beta +1)+(\alpha +1)(\beta -1)}{(\alpha +1)(\beta +1)}$

$=\frac{\mathit{\alpha \beta }+\alpha -\beta -1+\mathit{\alpha \beta }-\alpha +\beta -1}{\mathit{\alpha \beta }+\alpha +\beta +1}$

$=\frac{2\mathit{\alpha \beta }-2}{\mathit{\alpha \beta }+\alpha +\beta +1}$

$=\frac{6-2}{3+2+1}$

$=\frac{4}{6}$

$=\frac{2}{3}$
Product of roots, $\left(\frac{\alpha -1}{\alpha +1}\right)\left(\frac{\beta -1}{\beta +1}\right)$

$=\frac{\mathit{\alpha \beta }-\alpha -\beta +1}{\left(\mathit{\alpha \beta }+\alpha +\beta +1\right)}$

$=\frac{3-2+1}{3+2+1}$

$=\frac{2}{6}$

$=\frac{1}{3}$

Hence, equation, 

$x^2-(\mathrm{sum} \mathrm{of} \mathrm{roots})x+\left(\mathrm{product} \mathrm{of} \mathrm{roots}\right)=0$ 

$x^2-\left(\frac{2}{3}\right)x+\left(\frac{1}{3}\right)=0$ 

$3x^2-2x+1=0$