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(C) Given the equation $x^2 - 2x + 3 = 0$,the sum of roots is $\alpha + \beta = 2$ and the product of roots is $\alpha \beta = 3$.Let $y = \frac{x - 1

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the equation $x^2 - 2x + 3 = 0$,the sum of roots is $\alpha + \beta = 2$ and the product of roots is $\alpha \beta = 3$.Let $y = \frac{x - 1}{x + 1}$. Then $y(x + 1) = x - 1$,which implies $yx + y = x - 1$.Rearranging gives $x(y - 1) = -y - 1$,so $x = \frac{-(y + 1)}{y - 1} = \frac{y + 1}{1 - y}$.Substitute $x = \frac{y + 1}{1 - y}$ into the original equation $x^2 - 2x + 3 = 0$:$(\frac{y + 1}{1 - y})^2 - 2(\frac{y + 1}{1 - y}) + 3 = 0$.Multiply by $(1 - y)^2$: $(y + 1)^2 - 2(y + 1)(1 - y) + 3(1 - y)^2 = 0$.$(y^2 + 2y + 1) - 2(1 - y^2) + 3(1 - 2y + y^2) = 0$.$y^2 + 2y + 1 - 2 + 2y^2 + 3 - 6y + 3y^2 = 0$.Combining like terms: $(1 + 2 + 3)y^2 + (2 - 6)y + (1 - 2 + 3) = 0$.$6y^2 - 4y + 2 = 0$,which simplifies to $3y^2 - 2y + 1 = 0$.Thus,the required equation is $3x^2 - 2x + 1 = 0$.

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

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