If frac{alpha}{alpha+1} and frac{beta}{beta+1 | vedclass

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(A) Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$ $\Rightarrow yx + y = x$ $\Rightarrow x(y-1) = -y$ $\Rightarrow x = \frac{y}{1-y}$.Since $\frac{\alpha}

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

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(A) Let $y = \frac{x}{x+1}$. Then $y(x+1) = x$ $\Rightarrow yx + y = x$ $\Rightarrow x(y-1) = -y$ $\Rightarrow x = \frac{y}{1-y}$.Since $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ are roots of $x^2+7x+3=0$,substituting $x = \frac{y}{1-y}$ into the equation gives the equation for $y$ (which are $\alpha$ and $\beta$):$(\frac{y}{1-y})^2 + 7(\frac{y}{1-y}) + 3 = 0$Multiply by $(1-y)^2$:$y^2 + 7y(1-y) + 3(1-y)^2 = 0$$y^2 + 7y - 7y^2 + 3(1 - 2y + y^2) = 0$$y^2 + 7y - 7y^2 + 3 - 6y + 3y^2 = 0$$(1-7+3)y^2 + (7-6)y + 3 = 0$$-3y^2 + y + 3 = 0$$3y^2 - y - 3 = 0$.Thus,the equation with roots $\alpha$ and $\beta$ is $3x^2-x-3=0$.

If $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ are the roots of the quadratic equation $x^2+7x+3=0$,then the equation having roots $\alpha$ and $\beta$ is

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