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(A) Given the quadratic equation $px^2+qx-r=0$. Since $p, q, r$ are consecutive terms of a non-constant $G$.$P$.,we can write $q=pr$ and $r=pr^2$ (or

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$\frac{80}{9}$

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(A) Given the quadratic equation $px^2+qx-r=0$. Since $p, q, r$ are consecutive terms of a non-constant $G$.$P$.,we can write $q=pr$ and $r=pr^2$ (or using common ratio $k$,$q=pk, r=pk^2$).Substituting these into the equation: $px^2+pkx-pk^2=0$.Dividing by $p$ $(p 
eq 0)$: $x^2+kx-k^2=0$.From the roots $\alpha, \beta$,we have $\alpha+\beta = -k$ and $\alpha\beta = -k^2$.Given $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$,we have $\frac{\alpha+\beta}{\alpha\beta} = \frac{-k}{-k^2} = \frac{1}{k} = \frac{3}{4}$,so $k = \frac{4}{3}$.We need to find $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta$.$(\alpha-\beta)^2 = (-k)^2 - 4(-k^2) = k^2 + 4k^2 = 5k^2$.Substituting $k = \frac{4}{3}$: $(\alpha-\beta)^2 = 5 	imes (\frac{4}{3})^2 = 5 	imes \frac{16}{9} = \frac{80}{9}$.

Let $\alpha$ and $\beta$ be the roots of the equation $px^2+qx-r=0$,where $p \neq 0$. If $p, q,$ and $r$ are the consecutive terms of a non-constant $G$.$P$. and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$,then the value of $(\alpha-\beta)^2$ is:

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