If ac and B are the roots of the equation `4x^2-19x + 12=0`, find the equation having the roots `(1)/(alpha) and (1)/(beta)`

Options

A. `4x^(2) + 19x + 12=0`
B. `12x^(2)-19x + 4=0`
C. `12x^(2) + 19x + 4=0`
D. `4x^(2) + 19x-12=0`

Answer

To solve the problem, we need to find the equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \), given that \( \alpha \) and \( \beta \) are the roots of the equation \( 4x^2 - 19x + 12 = 0 \).

### Step-by-Step Solution:

1. **Identify the coefficients of the quadratic equation**:
The given equation is \( 4x^2 - 19x + 12 = 0 \).
Here, \( a = 4 \), \( b = -19 \), and \( c = 12 \).

2. **Calculate the sum and product of the roots**:
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{-19}{4} = \frac{19}{4} \).
- The product of the roots \( \alpha \beta = \frac{c}{a} = \frac{12}{4} = 3 \).

3. **Determine the new roots**:
We want to find the equation with roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).
- The sum of the new roots \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\frac{19}{4}}{3} = \frac{19}{12} \).
- The product of the new roots \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{3} \).

4. **Form the new quadratic equation**:
The general form of a quadratic equation with roots \( r_1 \) and \( r_2 \) is:
\[ x^2 - (r_1 + r_2)x + r_1 r_2 = 0 \]
Substituting the values we found:
\[ x^2 - \left(\frac{19}{12}\right)x + \frac{1}{3} = 0 \]

5. **Eliminate the fractions**:
To eliminate the fractions, multiply the entire equation by 12 (the least common multiple of the denominators):
\[ 12x^2 - 19x + 4 = 0 \]

### Final Answer:
The equation having the roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
\[ 12x^2 - 19x + 4 = 0 \]