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

Given: $\alpha \text{and} \beta$ be the roots of equation $p{x}^2+qx-r=0$

So, $\alpha +\beta =-\frac{q}{p}, \alpha \beta =-\frac{r}{p}$

It is given that, $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{3}{4}$

$\Rightarrow \frac{{\displaystyle \alpha +\beta }}{{\displaystyle \alpha \beta }}=\frac{{\displaystyle 3}}{{\displaystyle 4}}$

$\Rightarrow \frac{{\displaystyle q}}{{\displaystyle r}}=\frac{{\displaystyle 3}}{{\displaystyle 4}}$

Now, $p,q,r$ are in GP

So, common ratio of this GP will be $\frac{q}{p}=\frac{r}{q}=\frac{4}{3}$

$\Rightarrow p{x}^2+qx-r=0$

$\Rightarrow x^2+\frac{q}{p}x-\frac{r}{p}=0$

$\Rightarrow x^2+\frac{4}{3}x-{\left(\frac{4}{3}\right)}^2=0$

$\Rightarrow 9x^2+12x-16=0$

$\Rightarrow {\left(\alpha -\beta \right)}^2={\left(\alpha +\beta \right)}^2-4\alpha \beta$

$\Rightarrow {\left(\alpha -\beta \right)}^2={\left(\frac{-12}{9}\right)}^2-4\left(\frac{-16}{9}\right)$

$\Rightarrow {\left(\alpha -\beta \right)}^2=\frac{16}{9}+\frac{64}{9}$

$\Rightarrow {\left(\alpha -\beta \right)}^2=\frac{80}{9}$