If `alpha, beta` are the roots of `x ^(2) +x + 2 =0 and gamma, delta` are the roots of `x ^(2) + 3x + 4 =0.` then `(alpha + gamma) ( alpha +delta) ( beta + gamma )(beta+delta) ` is equal to

Options

A. `-18`
B. `18`
C. 24
D. 44

Answer

To solve the problem, we need to find the value of \((\alpha + \gamma)(\alpha + \delta)(\beta + \gamma)(\beta + \delta)\), where \(\alpha, \beta\) are the roots of the equation \(x^2 + x + 2 = 0\) and \(\gamma, \delta\) are the roots of the equation \(x^2 + 3x + 4 = 0\).

### Step 1: Find the roots of the first equation
The roots of the equation \(x^2 + x + 2 = 0\) can be calculated using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = 1\), and \(c = 2\).
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 8}}{2} = \frac{-1 \pm \sqrt{-7}}{2} = \frac{-1 \pm i\sqrt{7}}{2} \]
Thus, the roots are:
\[ \alpha = \frac{-1 + i\sqrt{7}}{2}, \quad \beta = \frac{-1 - i\sqrt{7}}{2} \]

### Step 2: Find the roots of the second equation
Now, we find the roots of the equation \(x^2 + 3x + 4 = 0\) using the same quadratic formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 16}}{2} = \frac{-3 \pm \sqrt{-7}}{2} = \frac{-3 \pm i\sqrt{7}}{2} \]
Thus, the roots are:
\[ \gamma = \frac{-3 + i\sqrt{7}}{2}, \quad \delta = \frac{-3 - i\sqrt{7}}{2} \]

### Step 3: Calculate \((\alpha + \gamma)(\alpha + \delta)(\beta + \gamma)(\beta + \delta)\)
We can express this product as:
\[ (\alpha + \gamma)(\alpha + \delta) = \alpha^2 + \alpha(\gamma + \delta) + \gamma\delta \]
\[ (\beta + \gamma)(\beta + \delta) = \beta^2 + \beta(\gamma + \delta) + \gamma\delta \]

### Step 4: Find \(\gamma + \delta\) and \(\gamma\delta\)
From the second equation \(x^2 + 3x + 4 = 0\):
- Sum of roots \(\gamma + \delta = -3\)
- Product of roots \(\gamma\delta = 4\)

### Step 5: Substitute values
Now substituting these values:
\[ (\alpha + \gamma)(\alpha + \delta) = \alpha^2 - 3\alpha + 4 \]
\[ (\beta + \gamma)(\beta + \delta) = \beta^2 - 3\beta + 4 \]

### Step 6: Calculate \(\alpha^2\) and \(\beta^2\)
Using the first equation \(x^2 + x + 2 = 0\):
- \(\alpha^2 = -\alpha - 2\)
- \(\beta^2 = -\beta - 2\)

### Step 7: Substitute \(\alpha^2\) and \(\beta^2\) into the expressions
Substituting \(\alpha^2\) and \(\beta^2\):
\[ (\alpha + \gamma)(\alpha + \delta) = (-\alpha - 2) - 3\alpha + 4 = -4\alpha + 2 \]
\[ (\beta + \gamma)(\beta + \delta) = (-\beta - 2) - 3\beta + 4 = -4\beta + 2 \]

### Step 8: Final multiplication
Now we multiply:
\[ (-4\alpha + 2)(-4\beta + 2) = 16\alpha\beta - 8(\alpha + \beta) + 4 \]
Using \(\alpha\beta = 2\) and \(\alpha + \beta = -1\):
\[ = 16(2) - 8(-1) + 4 = 32 + 8 + 4 = 44 \]

### Final Answer
Thus, the value of \((\alpha + \gamma)(\alpha + \delta)(\beta + \gamma)(\beta + \delta)\) is:
\[ \boxed{44} \]