If `alpha` and `beta` be the zeroes of the polynomial `p(x)=x^(2)-5x+2`, find the value of `(1)/(alpha)+(1)/(beta)-3alphabeta.`

Answer

To solve the problem, we need to find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} - 3\alpha\beta \) where \( \alpha \) and \( \beta \) are the zeroes of the polynomial \( p(x) = x^2 - 5x + 2 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial**:
The polynomial is given as \( p(x) = x^2 - 5x + 2 \).
Here, we can identify:
- \( a = 1 \)
- \( b = -5 \)
- \( c = 2 \)

2. **Calculate the sum and product of the roots**:
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5 \)
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{2}{1} = 2 \)

3. **Find \( \frac{1}{\alpha} + \frac{1}{\beta} \)**:
We can express \( \frac{1}{\alpha} + \frac{1}{\beta} \) as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta} \]
Substituting the values we found:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{5}{2} \]

4. **Calculate \( 3\alpha\beta \)**:
We already found \( \alpha \beta = 2 \), so:
\[ 3\alpha\beta = 3 \times 2 = 6 \]

5. **Combine the results**:
Now we substitute back into the expression we need to evaluate:
\[ \frac{1}{\alpha} + \frac{1}{\beta} - 3\alpha\beta = \frac{5}{2} - 6 \]
To perform the subtraction, convert 6 to a fraction with a denominator of 2:
\[ 6 = \frac{12}{2} \]
Therefore:
\[ \frac{5}{2} - \frac{12}{2} = \frac{5 - 12}{2} = \frac{-7}{2} \]

6. **Final Answer**:
The value of \( \frac{1}{\alpha} + \frac{1}{\beta} - 3\alpha\beta \) is:
\[ \boxed{-\frac{7}{2}} \]