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To find the quadratic equation with roots \( 7 + 5i \) and \( 7 - 5i \), we use the fact that if \(a\) and \(b\) are roots of a quadratic equation, then the equation can be written as:

\[
x^2 - (a+b)x + ab = 0
\]

Here:
- \( a = 7 + 5i \)
- \( b = 7 - 5i \)

1. **Calculate \(a + b\):**
   \[
   a + b = (7 + 5i) + (7 - 5i) = 14
   \]

2. **Calculate \(ab\):**
   \[
   ab = (7 + 5i)(7 - 5i) = 7^2 - (5i)^2 = 49 - (-25) = 49 + 25 = 74
   \]

3. **Form the quadratic equation:**
   Using \(a + b = 14\) and \(ab = 74\), the equation becomes:
   \[
   x^2 - 14x + 74 = 0
   \]

So, the correct answer is:
\[
\boxed{x^2 - 14x + 74 = 0}
\]

This corresponds to the last option in the choices provided.

Would you like more details on any step or have any questions?

Here are some related questions:
1. How do we derive a quadratic equation from its roots?
2. Why do complex roots always occur in conjugate pairs?
3. How do we verify the product of two complex conjugates?
4. How can you find the sum and product of roots in any quadratic equation?
5. What happens if a quadratic equation has real roots instead of complex roots?

**Tip:** Complex conjugates have the form \( a + bi \) and \( a - bi \); multiplying them always gives a real number.

Which quadratic equation has roots 7 + 5i and 7 - 5i?

Explore how to find the quadratic equation with complex roots 7 + 5i and 7 - 5i. This example walks through calculating the sum and product of roots, using concepts from algebra and complex numbers, suitable for students in Grades 10-12.

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