If `(2)/(3)` and `-(1)/(2)` are solution of quadratic equation `6x^(2)+ax-b=0`, find the values of a and b.

Answer

To find the values of \( a \) and \( b \) in the quadratic equation \( 6x^2 + ax - b = 0 \) given that the solutions are \( \frac{2}{3} \) and \( -\frac{1}{2} \), we can follow these steps:

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

### Step 2: Calculate the sum of the roots
The sum of the roots \( \alpha + \beta \) can be calculated as follows:
\[ \alpha + \beta = \frac{2}{3} + \left(-\frac{1}{2}\right) \]
To add these fractions, we need a common denominator. The least common multiple of 3 and 2 is 6.
\[ \alpha + \beta = \frac{2 \times 2}{6} + \left(-\frac{1 \times 3}{6}\right) = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

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

### Step 4: Form the quadratic equation using the roots
Using the sum and product of the roots, we can write the quadratic equation in the form:
\[ x^2 - (\alpha + \beta)x + \alpha \beta = 0 \]
Substituting the values we found:
\[ x^2 - \left(\frac{1}{6}\right)x - \left(-\frac{1}{3}\right) = 0 \]
This simplifies to:
\[ x^2 - \frac{1}{6}x + \frac{1}{3} = 0 \]

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

### Step 6: Compare with the original equation
Now we compare this equation \( 6x^2 - x + 2 = 0 \) with the given equation \( 6x^2 + ax - b = 0 \):
- From \( -x \), we have \( a = -1 \)
- From \( +2 \), we have \( b = 2 \)

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = -1, \quad b = 2 \]