If `alpha, beta` are the roots of `x^(2)+6x+9=0`, then the equation whose roots are `(1)/(alpha), (1)/(beta)` is

Options

A. `2x^(2)+3x-18=0`
B. `x^(2)+6x-9=0`
C. `9x^(2)+6x+1=0`
D. `9x^(2)-6x+1=0`

Answer

To solve the problem step by step, we will follow these instructions:

1. **Identify the given quadratic equation**:
The given equation is \( x^2 + 6x + 9 = 0 \).

2. **Find the roots of the equation**:
We can use the quadratic formula or factorization. The equation can be factored as:
\[ (x + 3)(x + 3) = 0 \]
Thus, the roots are:
\[ \alpha = -3, \quad \beta = -3 \]

3. **Calculate the sum and product of the roots**:
From Vieta's formulas, we know:
\[ \alpha + \beta = -b/a = -6/1 = -6 \]
\[ \alpha \beta = c/a = 9/1 = 9 \]

4. **Determine the new roots**:
We need to find the equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). Since \( \alpha = -3 \) and \( \beta = -3 \):
\[ \frac{1}{\alpha} = \frac{1}{-3} = -\frac{1}{3}, \quad \frac{1}{\beta} = \frac{1}{-3} = -\frac{1}{3} \]

5. **Find the sum and product of the new roots**:
The sum of the new roots is:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{1}{3} - \frac{1}{3} = -\frac{2}{3} \]
The product of the new roots is:
\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \left(-\frac{1}{3}\right) \cdot \left(-\frac{1}{3}\right) = \frac{1}{9} \]

6. **Form the new quadratic equation**:
The new quadratic equation can be formed using the sum and product of the roots:
\[ x^2 - \left(\text{sum of roots}\right)x + \left(\text{product of roots}\right) = 0 \]
Substituting the values:
\[ x^2 - \left(-\frac{2}{3}\right)x + \frac{1}{9} = 0 \]
This simplifies to:
\[ x^2 + \frac{2}{3}x + \frac{1}{9} = 0 \]

7. **Clear the fractions**:
To eliminate the fractions, multiply the entire equation by 9:
\[ 9x^2 + 6x + 1 = 0 \]

8. **Final answer**:
The equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
\[ 9x^2 + 6x + 1 = 0 \]