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

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To solve the problem, let's use Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots.

Given the quadratic polynomial:
$$ f(x) = x^2 - x - 4 $$
Let $$ \alpha $$ and $$ \beta $$ be the roots of this polynomial. By Vieta's formulas:
$$
\alpha + \beta = 1 \quad \text{(sum of roots)}
$$
$$
\alpha \beta = -4 \quad \text{(product of roots)}
$$

We need to find the value of:
$$
\frac{1}{\alpha} + \frac{1}{\beta} - \alpha \beta
$$

First, compute $$ \frac{1}{\alpha} + \frac{1}{\beta} $$:
$$
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}
$$
Substituting the known values:
$$
\frac{\alpha + \beta}{\alpha \beta} = \frac{1}{-4} = -\frac{1}{4}
$$

Now, subtract $$ \alpha \beta $$ from this result:
$$
-\frac{1}{4} - \alpha \beta = -\frac{1}{4} - (-4) = -\frac{1}{4} + 4 = \frac{15}{4}
$$

**Answer:** $$\boxed{\dfrac{15}{4}}$$
The value of $$ \frac{1}{\alpha} + \frac{1}{\beta} - \alpha \beta $$ is $$ \frac{15}{4} $$.

If and are the zeroes of the quadratic polynomial
, find the value of .

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