Find the quadratic equation whose roots are `2/3` and `-1/2`:

Options

A. `6x^2+x+2=0`
B. `6x^2-x-2=0`
C. `16x^2-x+2=0`
D. `16x^2+x-2=0`

Answer

To find the quadratic equation whose roots are \( \frac{2}{3} \) and \( -\frac{1}{2} \), we will follow these steps:

### Step 1: Identify the roots
The roots given are:
- \( r_1 = \frac{2}{3} \)
- \( r_2 = -\frac{1}{2} \)

### Step 2: Calculate the sum of the roots
The sum of the roots \( S \) is given by:
\[ S = r_1 + r_2 = \frac{2}{3} + \left(-\frac{1}{2}\right) \]
To add these fractions, we need a common denominator. The least common multiple of 3 and 2 is 6.
\[ S = \frac{2}{3} = \frac{4}{6} \quad \text{and} \quad -\frac{1}{2} = -\frac{3}{6} \]
Thus,
\[ S = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

### Step 3: Calculate the product of the roots
The product of the roots \( P \) is given by:
\[ P = r_1 \cdot r_2 = \frac{2}{3} \cdot \left(-\frac{1}{2}\right) = -\frac{2}{6} = -\frac{1}{3} \]

### Step 4: Form the quadratic equation
The standard form of a quadratic equation with roots \( r_1 \) and \( r_2 \) is:
\[ x^2 - Sx + P = 0 \]
Substituting the values of \( S \) and \( P \):
\[ x^2 - \frac{1}{6}x - \frac{1}{3} = 0 \]

### Step 5: Eliminate the fractions
To eliminate the fractions, multiply the entire equation by 6 (the least common multiple of the denominators):
\[ 6x^2 - x - 2 = 0 \]

### Conclusion
The quadratic equation whose roots are \( \frac{2}{3} \) and \( -\frac{1}{2} \) is:
\[ 6x^2 - x - 2 = 0 \]