If `alpha and beta` are the roots of the equation `x^(2)-3x+2=0,` Find the quadratic equation whose roots are `-alpha and -beta`:

Options

A. `x^(2)-3x+2=0`
B. `x^(2)+3x+2=0`
C. `x^(2)+3x-2=0`
D. none of these

Answer

To find the quadratic equation whose roots are \(-\alpha\) and \(-\beta\), where \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 - 3x + 2 = 0\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the coefficients of the given quadratic equation**:
The given equation is \(x^2 - 3x + 2 = 0\). Here, we can identify:
- \(a = 1\)
- \(b = -3\)
- \(c = 2\)

2. **Calculate the sum and product of the roots**:
Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-3}{1} = 3\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{2}{1} = 2\)

3. **Determine the new roots**:
The new roots we want are \(-\alpha\) and \(-\beta\).

4. **Calculate the sum and product of the new roots**:
- The sum of the new roots \(-\alpha - \beta = -(\alpha + \beta) = -3\)
- The product of the new roots \((- \alpha)(- \beta) = \alpha \beta = 2\)

5. **Form the new quadratic equation**:
The standard form of a quadratic equation with roots \(r_1\) and \(r_2\) is given by:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
Substituting the values of the new roots:
\[ x^2 - (-3)x + 2 = 0 \]
This simplifies to:
\[ x^2 + 3x + 2 = 0 \]

### Final Answer:
The quadratic equation whose roots are \(-\alpha\) and \(-\beta\) is:
\[ x^2 + 3x + 2 = 0 \]