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If $\alpha, \beta$ are the zeroes of the polynomial $f(x)=x^{2}+x-2$, then $\left(\frac{\alpha}{\beta}-\frac{\alpha}{\beta}\right)$.
If $\alpha, \beta$ are the zeroes of the polynomial $f(x)=x^{2}+x-2$, then $\left(\frac{\alpha}{\beta}-\frac{\alpha}{\beta}\right)$.
CBSE | Class 10 | Excercise 2C | Maths | Polynomials | RS Aggarwal
If $\alpha, \beta$ are the zeroes of the polynomial $f(x)=x^{2}+x-2$, then $\left(\frac{\alpha}{\beta}-\frac{\alpha}{\beta}\right)$.

using the relationship between the zeroes of the quadratic polynomial. We have

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

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

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

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

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

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

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

$=\frac{9}{4}$

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

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

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