 # If α and β are the zeroes of a polynomial Question:

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

Solution:

By using the relationship between the zeroes of the quadratic ploynomial.
We have,

Sum of zeroes $=\frac{-(\text { coefficient of } x)}{\text { coefficent of } x^{2}}$ and Product of zeroes $=\frac{\text { constant term }}{\text { coefficent of } x^{2}}$

$\Rightarrow \alpha+\beta=\frac{-1}{1}$ and $\alpha \beta=\frac{-2}{1}$

$\Rightarrow \alpha+\beta=-1$ and $\alpha \beta=-2$

Now, $\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\left(\frac{\beta-\alpha}{\alpha \beta}\right)^{2}$

$=\frac{(\alpha+\beta)^{2}-4 \alpha \beta}{(\alpha \beta)^{2}} \quad\left[\because(\beta-\alpha)^{2}=(\alpha+\beta)^{2}-4 \alpha \beta\right]$

$=\frac{(-1)^{2}-4(-2)}{(-2)^{2}} \quad[\because \alpha+\beta=-1$ and $\alpha \beta=-2]$

$=\frac{(-1)^{2}-4(-2)}{4}$

$=\frac{9}{4}$

$\because\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\frac{9}{4}$

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