If α and β are the zeros of the quadratic polynomial $f\left(x\right)=x^2+x-2$, then find the value of $\frac{1}{\alpha }-\frac{1}{\beta }$.

Given α and β are the zeros of the quadratic polynomial $f\left(x\right)=x^2+x-2$

$\mathrm{Sum} \mathrm{of} \mathrm{zeroes}=-\frac{\mathrm{b}}{\mathrm{a}}$

$\Rightarrow \alpha +\beta =-1$     -----(1) 

$\mathrm{Product} \mathrm{of} \mathrm{zeroes}=\frac{c}{\mathrm{a}}$

$\Rightarrow \alpha \beta =-2$     -----(2)

Consider, $\frac{1}{\alpha }-\frac{1}{\beta }$$=\frac{\beta -\alpha }{\mathit{\alpha \beta }}$

$\frac{1}{\alpha }-\frac{1}{\beta }$$=\frac{-(\alpha -\beta )}{\mathit{\alpha \beta }}$     -----(3)

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

$\Rightarrow \alpha -\beta =\sqrt{{\left(-1\right)}^2-4(-2)}$

$\Rightarrow \alpha -\beta =\sqrt{1+8}=\sqrt{9}$

$\Rightarrow \alpha -\beta =3$      -----(4)

Putting the value of αβ and α−β in equation (3)

$\frac{1}{\alpha }-\frac{1}{\beta }=\frac{-\left(3\right)}{-2}$

$\frac{1}{\alpha }-\frac{1}{\beta }=\frac{3}{2}$

Hence, the value of $\frac{1}{\alpha }-\frac{1}{\beta }$ is $\frac{3}{2}$.