If `alpha` and `beta` are zeroes of `8x^(2)-6x+1`, then find the value of `(1)/(alpha)+(1)/(beta)`.

Options

A. 6
B. 8
C. `-6`
D. `-8`

Explanation

To find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) where \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(8x^2 - 6x + 1\), we can follow these steps:

### Step 1: Identify coefficients
The given polynomial is \(8x^2 - 6x + 1\). Here, we identify:
- \(a = 8\)
- \(b = -6\)
- \(c = 1\)

### Step 2: Use the relationships for the roots
From Vieta's formulas, we know:
- The sum of the roots (\(\alpha + \beta\)) is given by \(-\frac{b}{a}\).
- The product of the roots (\(\alpha \cdot \beta\)) is given by \(\frac{c}{a}\).

### Step 3: Calculate \(\alpha + \beta\)
Using the formula for the sum of the roots:
\[ \alpha + \beta = -\frac{-6}{8} = \frac{6}{8} = \frac{3}{4} \]

### Step 4: Calculate \(\alpha \cdot \beta\)
Using the formula for the product of the roots:
\[ \alpha \cdot \beta = \frac{1}{8} \]

### Step 5: Find \(\frac{1}{\alpha} + \frac{1}{\beta}\)
We can express \(\frac{1}{\alpha} + \frac{1}{\beta}\) in terms of \(\alpha + \beta\) and \(\alpha \cdot \beta\):
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta} \]

### Step 6: Substitute the values
Substituting the values we found:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{3}{4}}{\frac{1}{8}} \]

### Step 7: Simplify the expression
To simplify:
\[ \frac{3}{4} \div \frac{1}{8} = \frac{3}{4} \times 8 = \frac{3 \times 8}{4} = \frac{24}{4} = 6 \]

### Final Answer
Thus, the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is \(6\).

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