Form the quadratic equation, if the roots are
`-(1)/(2)and(1)/(3)`

Answer

To form a quadratic equation given the roots \(-\frac{1}{2}\) and \(\frac{1}{3}\), we can follow these steps:

### Step 1: Identify the roots
Let the roots be \(\alpha = -\frac{1}{2}\) and \(\beta = \frac{1}{3}\).

### Step 2: Calculate the sum of the roots
The sum of the roots \(\alpha + \beta\) is calculated as follows:
\[ \alpha + \beta = -\frac{1}{2} + \frac{1}{3} \]
To add these fractions, we need a common denominator. The least common multiple (LCM) of 2 and 3 is 6.
\[ \alpha + \beta = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6} \]

### Step 3: Calculate the product of the roots
The product of the roots \(\alpha \beta\) is calculated as follows:
\[ \alpha \beta = \left(-\frac{1}{2}\right) \times \left(\frac{1}{3}\right) = -\frac{1}{6} \]

### Step 4: Form the quadratic equation
Using the standard form of a quadratic equation \(x^2 - (\alpha + \beta)x + \alpha \beta = 0\), we substitute the values we calculated:
\[ x^2 - \left(-\frac{1}{6}\right)x - \frac{1}{6} = 0 \]
This simplifies to:
\[ x^2 + \frac{1}{6}x - \frac{1}{6} = 0 \]

### Step 5: Eliminate the fraction by multiplying through by 6
To eliminate the fractions, multiply the entire equation by 6:
\[ 6x^2 + x - 1 = 0 \]

### Final Answer
The quadratic equation formed is:
\[ 6x^2 + x - 1 = 0 \]

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