Form a quadratic polynomial whose zeroes are

`(3)/(5) and -(1)/(2)`

Options

A. `x^(2) - 9x + 6`
B. `10x^(2) - x - 3`
C. `9x^(2) + x + 6`
D. `7x^(2) - 3x + 4`

Answer

To form a quadratic polynomial whose zeroes are \( \frac{3}{5} \) and \( -\frac{1}{2} \), we can follow these steps:

### Step 1: Identify the zeroes
Let the zeroes (roots) of the polynomial be:
- \( \alpha = \frac{3}{5} \)
- \( \beta = -\frac{1}{2} \)

### Step 2: Calculate the sum of the zeroes
The sum of the zeroes \( \alpha + \beta \) is calculated as follows:
\[ \alpha + \beta = \frac{3}{5} + \left(-\frac{1}{2}\right) \]
To add these fractions, we need a common denominator. The least common multiple (LCM) of 5 and 2 is 10.

Converting both fractions:
\[ \frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10} \]
\[ -\frac{1}{2} = -\frac{1 \times 5}{2 \times 5} = -\frac{5}{10} \]
Now, adding them together:
\[ \alpha + \beta = \frac{6}{10} - \frac{5}{10} = \frac{1}{10} \]

### Step 3: Calculate the product of the zeroes
The product of the zeroes \( \alpha \beta \) is calculated as follows:
\[ \alpha \beta = \frac{3}{5} \times -\frac{1}{2} = -\frac{3}{10} \]

### Step 4: Form the quadratic polynomial
Using the standard form of a quadratic polynomial given the zeroes:
\[ f(x) = x^2 - (\alpha + \beta)x + \alpha \beta \]
Substituting the values we calculated:
\[ f(x) = x^2 - \left(\frac{1}{10}\right)x - \frac{3}{10} \]

### Step 5: Eliminate the fractions
To eliminate the fractions, we can multiply the entire polynomial by 10:
\[ 10f(x) = 10\left(x^2 - \frac{1}{10}x - \frac{3}{10}\right) \]
This simplifies to:
\[ 10f(x) = 10x^2 - x - 3 \]

### Final Polynomial
Thus, the quadratic polynomial is:
\[ f(x) = 10x^2 - x - 3 \]