Question

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

Answer 1

Given if $\alpha$ and $\beta$​​​​​ are the solutions of the polynomial $f\left(x\right)=x^2+x-2$.
So, first let us find zeros of $f\left(x\right)=0$:
The middle term $x$ is expressed as sum of $2x$ and $-x$ such that its product is equals to product of extreme terms.
$(-2)\times x^2=-2x^2$
Thus, $x^2+2x-x-2=0$
$x(x+2)-1(x+2)=0$
$(x+2)(x-1)=0$
$(x+2)=0\text{or}(x-1)=0$
$\Rightarrow x=-2\text{or}x=1$
$\therefore \alpha ,\beta =(1,-2)\text{or}(-2,1)$
Case(i): when $(\alpha ,\beta )=(1,-2)$
$(\frac{1}{\alpha }-\frac{1}{\beta })=\frac{1}{1}-\frac{1}{-2}$
$=1+\frac{1}{2}$
$=\frac{2+1}{2}$
$\therefore \frac{1}{\alpha }-\frac{1}{\beta }=\frac{3}{2}$
Case(ii): When $(\alpha ,\beta )=(-2,1))$
Consider $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{1}{-2}-\frac{1}{1}$
$=\frac{-1-2}{2}$
$\therefore \frac{1}{\alpha }-\frac{1}{\beta }=\frac{-3}{2}$
Hence, $\frac{1}{\alpha }-\frac{1}{\beta }=\frac{-3}{2}$ or $\frac{3}{2}$