If `alpha` `beta` are the zeroes of the polynomial f(x) = `x^2 - 7x + 12`, then find the value of `1/(alpha) + 1/beta`

Options

A. 12
B. `-7/12`
C. `-7`
D. `7/12`

Answer

To solve the problem, we need to find the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) where \( \alpha \) and \( \beta \) are the roots of the polynomial \( f(x) = x^2 - 7x + 12 \).

### Step 1: Find the roots of the polynomial
The polynomial can be factored or solved using the quadratic formula. Here, we will factor it.

The polynomial is:
\[ f(x) = x^2 - 7x + 12 \]

We look for two numbers that multiply to \( 12 \) (the constant term) and add up to \( -7 \) (the coefficient of \( x \)). The numbers \( -3 \) and \( -4 \) satisfy this condition because:
\[ -3 \times -4 = 12 \quad \text{and} \quad -3 + -4 = -7 \]

Thus, we can factor the polynomial as:
\[ f(x) = (x - 3)(x - 4) \]

### Step 2: Identify the roots
Setting \( f(x) = 0 \):
\[ (x - 3)(x - 4) = 0 \]
This gives us the roots:
\[ x = 3 \quad \text{and} \quad x = 4 \]
So, we have \( \alpha = 3 \) and \( \beta = 4 \).

### Step 3: Calculate \( \frac{1}{\alpha} + \frac{1}{\beta} \)
Now we need to find:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{3} + \frac{1}{4} \]

To add these fractions, we need a common denominator. The least common multiple of \( 3 \) and \( 4 \) is \( 12 \).

### Step 4: Find a common denominator and add
Rewriting the fractions:
\[ \frac{1}{3} = \frac{4}{12} \quad \text{and} \quad \frac{1}{4} = \frac{3}{12} \]
Now, we can add them:
\[ \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} \]

### Final Answer
Thus, the value of \( \frac{1}{\alpha} + \frac{1}{\beta} \) is:
\[ \frac{7}{12} \]