If roots of the equation `3x^2-6x+5=0` are oc and B, find the equation having the roots `(1)/(alpha) and (1)/(beta)`

Answer

To solve the problem, we need to find the equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \), where \( \alpha \) and \( \beta \) are the roots of the given quadratic equation \( 3x^2 - 6x + 5 = 0 \).

### Step 1: Identify the coefficients
The given quadratic equation is:
\[ 3x^2 - 6x + 5 = 0 \]
Here, the coefficients are:
- \( A = 3 \)
- \( B = -6 \)
- \( C = 5 \)

### Step 2: Calculate the sum and product of the roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{B}{A} = -\frac{-6}{3} = \frac{6}{3} = 2 \]

- The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{C}{A} = \frac{5}{3} \]

### Step 3: Find the sum and product of the new roots
The new roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). We need to find their sum and product:
- The sum of the new roots \( \frac{1}{\alpha} + \frac{1}{\beta} \) can be calculated as:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{2}{\frac{5}{3}} = \frac{2 \times 3}{5} = \frac{6}{5} \]

- The product of the new roots \( \frac{1}{\alpha} \cdot \frac{1}{\beta} \) is:
\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \]

### Step 4: Form the new quadratic equation
Using the sum and product of the new roots, we can form the new quadratic equation:
\[ x^2 - \left(\text{sum of roots}\right)x + \left(\text{product of roots}\right) = 0 \]
Substituting the values we found:
\[ x^2 - \frac{6}{5}x + \frac{3}{5} = 0 \]

### Step 5: Clear the fractions
To eliminate the fractions, we can multiply the entire equation by 5:
\[ 5x^2 - 6x + 3 = 0 \]

### Final Answer
The equation having the roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:
\[ 5x^2 - 6x + 3 = 0 \]