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

From the equation x^2 - 2x + 3 = 0, the sum of the roots is alpha + beta = 2 and the product is alpha * beta = 3. Let the new root be y = (x - 1)/(x + 1), which rearranges to x = (1 + y)/(1 - y). Substituting this expression for x into the original quadratic equation yields ((1 + y)/(1 - y))^2 - 2((1 + y)/(1 - y)) + 3 = 0. Clearing the denominator by multiplying by (1 - y)^2 and simplifying the algebraic expansion gives the new equation 3y^2 - 2y + 1 = 0.

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