If the sum of square of roots of equation `x^2+(p+iq)x+3i=0` is 8, then find p and q, where p and q are real.

Options

A. `p=2,q=2`
B. `p=-3,q=4`
C. `p=3,q=1 or p=-3,q=-1`
D. `p=-2,q=-2`

Explanation

To solve the problem, we need to find the values of \( p \) and \( q \) such that the sum of the squares of the roots of the equation \( x^2 + (p + iq)x + 3i = 0 \) is equal to 8.

### Step-by-step Solution:

1. **Identify the Roots and Their Properties**:
The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be denoted as \( \alpha \) and \( \beta \). The sum of the roots \( \alpha + \beta \) is given by \( -\frac{b}{a} \) and the product of the roots \( \alpha \beta \) is given by \( \frac{c}{a} \).

For our equation:
- \( a = 1 \)
- \( b = p + iq \)
- \( c = 3i \)

Thus:
\[ \alpha + \beta = -\frac{p + iq}{1} = -p - iq \]
\[ \alpha \beta = \frac{3i}{1} = 3i \]

2. **Use the Given Condition**:
We know that the sum of the squares of the roots can be expressed as:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Setting this equal to 8 gives us:
\[ (\alpha + \beta)^2 - 2\alpha\beta = 8 \]

3. **Substituting the Values**:
Substitute the expressions for \( \alpha + \beta \) and \( \alpha \beta \):
\[ (-p - iq)^2 - 2(3i) = 8 \]

4. **Expand the Square**:
Expanding \( (-p - iq)^2 \):
\[ (-p - iq)^2 = p^2 + 2pq i - q^2 \]
Therefore, we have:
\[ (p^2 - q^2 + 2pq i) - 6i = 8 \]

5. **Separating Real and Imaginary Parts**:
Now, equate the real and imaginary parts:
- Real part: \( p^2 - q^2 = 8 \)
- Imaginary part: \( 2pq - 6 = 0 \)

6. **Solving the Equations**:
From the imaginary part equation:
\[ 2pq = 6 \implies pq = 3 \]
Now we have a system of equations:
1. \( p^2 - q^2 = 8 \)
2. \( pq = 3 \)

7. **Express \( q \) in Terms of \( p \)**:
From \( pq = 3 \), we can express \( q \):
\[ q = \frac{3}{p} \]

8. **Substituting \( q \) into the First Equation**:
Substitute \( q \) into the first equation:
\[ p^2 - \left(\frac{3}{p}\right)^2 = 8 \]
Simplifying gives:
\[ p^2 - \frac{9}{p^2} = 8 \]
Multiplying through by \( p^2 \) to eliminate the fraction:
\[ p^4 - 8p^2 - 9 = 0 \]

9. **Letting \( x = p^2 \)**:
Let \( x = p^2 \):
\[ x^2 - 8x - 9 = 0 \]

10. **Using the Quadratic Formula**:
Solving this quadratic using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm 10}{2} \]
This gives us:
\[ x = 9 \quad \text{or} \quad x = -1 \]
Since \( x = p^2 \), we discard \( x = -1 \).

11. **Finding \( p \)**:
Therefore, \( p^2 = 9 \) implies:
\[ p = 3 \quad \text{or} \quad p = -3 \]

12. **Finding Corresponding \( q \)**:
For \( p = 3 \):
\[ q = \frac{3}{3} = 1 \]
For \( p = -3 \):
\[ q = \frac{3}{-3} = -1 \]

### Final Answers:
Thus, the pairs \( (p, q) \) that satisfy the conditions are \( (3, 1) \) and \( (-3, -1) \).