If α and β are the zeros of the quadratic polynomial f(x)

Solution:

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomials $f(x)=x^{2}-x-4$

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

$\alpha \beta=\frac{-4}{1}$

$\alpha \beta=-4$

We have,

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

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

By substituting $\alpha+\beta=1$ and $\alpha \beta=-4$ we get,

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{1}{-4}-(-4)$

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{1}{-4}+\frac{4}{1}$

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{1}{-4}+\frac{4 \times 4}{1 \times 4}$

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{1}{-4}+\frac{16}{4}$

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{-1+16}{4}$

$\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta=\frac{15}{4}$

Hence, the value of $\frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta$ is $\frac{15}{4}$.