What is the sum of the reciprocals of the values of zeroes of the polynomial `6x^(2) +3x^(2)-5x+1`?

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the sum of the reciprocals of the values of the zeroes of the polynomial \(6x^2 + 3x^2 - 5x + 1\), we can follow these steps:

### Step 1: Combine Like Terms
First, we simplify the polynomial by combining like terms:
\[ 6x^2 + 3x^2 = 9x^2 \]
Thus, the polynomial can be rewritten as:
\[ 9x^2 - 5x + 1 \]

### Step 2: Identify Coefficients
Next, we identify the coefficients \(a\), \(b\), and \(c\) from the polynomial \(ax^2 + bx + c\):
- \(a = 9\)
- \(b = -5\)
- \(c = 1\)

### Step 3: Use Vieta's Formulas
According to Vieta's formulas, for a quadratic polynomial \(ax^2 + bx + c\):
- The sum of the roots (\(\alpha + \beta\)) is given by:
\[ \alpha + \beta = -\frac{b}{a} \]
- The product of the roots (\(\alpha \beta\)) is given by:
\[ \alpha \beta = \frac{c}{a} \]

### Step 4: Calculate the Sum and Product of Roots
Now, we can calculate the sum and product of the roots using the coefficients:
1. **Sum of the roots**:
\[ \alpha + \beta = -\frac{-5}{9} = \frac{5}{9} \]
2. **Product of the roots**:
\[ \alpha \beta = \frac{1}{9} \]

### Step 5: Find the Sum of the Reciprocals of the Roots
The sum of the reciprocals of the roots (\(\frac{1}{\alpha} + \frac{1}{\beta}\)) can be expressed as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \]
Substituting the values we found:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\frac{5}{9}}{\frac{1}{9}} = \frac{5}{9} \times 9 = 5 \]

### Final Answer
Thus, the sum of the reciprocals of the values of the zeroes of the polynomial is:
\[ \boxed{5} \]