If α and β are the zeros of the quadratic polynomial

Question:

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

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomials $f(x)=x^{2}+x-2$

Sum of the zeros $=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$

$\alpha+\beta=-\left[\frac{1}{1}\right]$

$\alpha+\beta=-\frac{1}{1}$

$\alpha+\beta=-1$

Product if zeros $=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

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

$\alpha \beta=-2$

We have, $\frac{1}{\alpha}-\frac{1}{\beta}$

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

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

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

By substituting $\alpha+\beta=-1$ and $\alpha \beta=-2$ we get,

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

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

$\frac{1}{4}+\frac{1 \times 4}{1 \times 4}=\left(\frac{1}{\alpha}\right)^{2}+\left(\frac{1}{\beta}\right)^{2}$

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

$\frac{5}{4}=\left(\frac{1}{\alpha}\right)^{2}+\left(\frac{1}{\beta}\right)^{2}$

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

By substituting $\frac{5}{4}=\left(\frac{1}{\alpha}\right)^{2}+\left(\frac{1}{\beta}\right)^{2}$ in $\left(\frac{1}{\alpha}\right)^{2}+\left(\frac{1}{\beta}\right)^{2}-\frac{2}{\alpha \beta}$ we get,

$\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\frac{5}{4}-\frac{2}{-2}$

$\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\frac{5}{4}+1$

$\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\frac{5}{4}+\frac{1 \times 4}{1 \times 4}$

$\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}=\frac{5+4}{4}$

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

Taking square root on both sides we get

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

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

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

 

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