34. If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial \( f(x)=x^{2}-x-4 \), find the value of \( \frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta \).

To find the value of \( \frac{1}{\alpha}+\frac{1}{\beta}-\alpha \beta \) where \( \alpha \) and \( \beta \) are the roots of the polynomial \( f(x) = x^2 - x - 4 \), we can use Vieta's formulas: 1. \( \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = 1 \) 2. \( \alpha \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = -4 \) Now, we calculate \( \frac{1}{\alpha} + \frac{1}{\beta} \): \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \] Substituting the values from Vieta's formulas: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{-4} = -\frac{1}{4} \] Next, we substitute \( \alpha \beta \) into the expression: \[ \frac{1}{\alpha} + \frac{1}{\beta} - \alpha \beta = -\frac{1}{4} - (-4) = -\frac{1}{4} + 4 \] To combine these, convert \( 4 \) to a fraction: \[ 4 = \frac{16}{4} \Rightarrow -\frac{1}{4} + \frac{16}{4} = \frac{15}{4} \] Thus, the final answer is: \[ \frac{1}{\alpha} + \frac{1}{\beta} - \alpha \beta = \frac{15}{4} \]