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(A) Given the equation $x^2 - 5x - 3 = 0$,we have $\alpha + \beta = 5$ and $\alpha\beta = -3$.Let $y = \frac{1}{2\alpha - 3}$. Then $2\alpha - 3 = \fr

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$33x^2 + 4x - 1 = 0$

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(A) Given the equation $x^2 - 5x - 3 = 0$,we have $\alpha + \beta = 5$ and $\alpha\beta = -3$.Let $y = \frac{1}{2\alpha - 3}$. Then $2\alpha - 3 = \frac{1}{y}$,so $2\alpha = \frac{1}{y} + 3 = \frac{1 + 3y}{y}$,which gives $\alpha = \frac{3y + 1}{2y}$.Since $\alpha$ is a root of $x^2 - 5x - 3 = 0$,we substitute $\alpha$ into the equation:$(\frac{3y + 1}{2y})^2 - 5(\frac{3y + 1}{2y}) - 3 = 0$.Multiply by $4y^2$ to clear the denominator:$(3y + 1)^2 - 10y(3y + 1) - 12y^2 = 0$.$9y^2 + 6y + 1 - 30y^2 - 10y - 12y^2 = 0$.$-33y^2 - 4y + 1 = 0$.Multiplying by $-1$,we get $33y^2 + 4y - 1 = 0$.Replacing $y$ with $x$,the required equation is $33x^2 + 4x - 1 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 5x - 3 = 0$,then what is the equation whose roots are $\frac{1}{2\alpha - 3}$ and $\frac{1}{2\beta - 3}$?

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