Find an equation, whose roots are `(1)/(2)` and `(1)/(3)` / ऐसा समीकरण ज्ञात कीजिए जिसके मूल `(1)/(2)` तथा `(1)/(3)` हैं -

Options

A. `6x^2 - 5x+1=0`
B. `2x^2 - 4x +9 = 0`
C. `4x^2 - x - 18 = 0`
D. `4x^2 - x - 18 = 0`

Explanation

To find an equation whose roots are \( \frac{1}{2} \) and \( \frac{1}{3} \), we can use the standard form of a quadratic equation based on its roots. The general formula for a quadratic equation with roots \( r_1 \) and \( r_2 \) is:

\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]

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

### Step 2: Calculate the sum of the roots
The sum of the roots \( r_1 + r_2 \) is calculated as follows:

\[ r_1 + r_2 = \frac{1}{2} + \frac{1}{3} \]

To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6.

\[ r_1 + r_2 = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \]

### Step 3: Calculate the product of the roots
The product of the roots \( r_1 \cdot r_2 \) is calculated as follows:

\[ r_1 \cdot r_2 = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]

### Step 4: Substitute the values into the quadratic equation
Now, we substitute the sum and product of the roots into the quadratic equation formula:

\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]

Substituting the values we calculated:

\[ x^2 - \left(\frac{5}{6}\right)x + \frac{1}{6} = 0 \]

### Step 5: Eliminate fractions by multiplying through by the least common multiple
To eliminate the fractions, we can multiply the entire equation by 6 (the least common multiple of the denominators):

\[ 6 \cdot x^2 - 6 \cdot \left(\frac{5}{6}\right)x + 6 \cdot \frac{1}{6} = 0 \]

This simplifies to:

\[ 6x^2 - 5x + 1 = 0 \]

### Final Equation
Thus, the equation whose roots are \( \frac{1}{2} \) and \( \frac{1}{3} \) is:

\[ 6x^2 - 5x + 1 = 0 \]