If `alpha and beta` are the roots of equation `2x^(2)-x-3=0`, then find a equation having roots `(2alpha+3)and(2beta+3)`.

Answer

To solve the problem, we need to find the equation whose roots are \(2\alpha + 3\) and \(2\beta + 3\), given that \(\alpha\) and \(\beta\) are the roots of the equation \(2x^2 - x - 3 = 0\).

### Step 1: Find the roots \(\alpha\) and \(\beta\)

We start with the quadratic equation:

\[ 2x^2 - x - 3 = 0 \]

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \(a = 2\), \(b = -1\), and \(c = -3\).

Substituting the values into the formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \]

Calculating the discriminant:

\[ x = \frac{1 \pm \sqrt{1 + 24}}{4} \]
\[ x = \frac{1 \pm \sqrt{25}}{4} \]
\[ x = \frac{1 \pm 5}{4} \]

This gives us two roots:

\[ \alpha = \frac{6}{4} = \frac{3}{2}, \quad \beta = \frac{-4}{4} = -1 \]

### Step 2: Find \(2\alpha + 3\) and \(2\beta + 3\)

Now we calculate the new roots:

1. For \(2\alpha + 3\):

\[ 2\alpha + 3 = 2 \cdot \frac{3}{2} + 3 = 3 + 3 = 6 \]

2. For \(2\beta + 3\):

\[ 2\beta + 3 = 2 \cdot (-1) + 3 = -2 + 3 = 1 \]

### Step 3: Form the new quadratic equation

Now we need to find the quadratic equation whose roots are \(6\) and \(1\). The general form of a quadratic equation with roots \(p\) and \(q\) is:

\[ x^2 - (p + q)x + pq = 0 \]

Here, \(p = 6\) and \(q = 1\):

1. Calculate \(p + q\):

\[ p + q = 6 + 1 = 7 \]

2. Calculate \(pq\):

\[ pq = 6 \cdot 1 = 6 \]

Thus, the equation is:

\[ x^2 - 7x + 6 = 0 \]

### Final Answer

The required equation having roots \(2\alpha + 3\) and \(2\beta + 3\) is:

\[ \boxed{x^2 - 7x + 6 = 0} \]