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Question 12 asks:

"Granted that a certain quartic equation has the roots \( 2 \pm 3i \), and no imaginary roots different from them, does it have two real roots?"

### Solution:

Given roots \( 2 \pm 3i \), these are a pair of complex conjugates. Since the equation is a quartic (degree 4), it must have 4 roots in total. The question also specifies that there are no imaginary roots different from \( 2 \pm 3i \), which means these two complex roots are the only imaginary roots.

Now, to satisfy the condition of a quartic equation with 4 roots, the remaining two roots must be real. Thus, **the equation has two real roots** in addition to the two imaginary roots \( 2 \pm 3i \).

### Conclusion:
Yes, the quartic equation has two real roots.

Would you like further details or clarifications?

Here are five related questions to extend this concept:
1. What is the general form of a quartic equation with complex conjugate roots?
2. How does the Fundamental Theorem of Algebra help in identifying the number of roots for polynomials?
3. Can a quartic equation have four real roots? If so, under what conditions?
4. How do the coefficients of a polynomial affect whether the roots are real or complex?
5. What is the relationship between complex conjugate pairs and the discriminant of a polynomial?

### Tip:
For polynomials with real coefficients, complex roots always appear in conjugate pairs.

Granted that a certain quartic equation has the roots 2 ± 3i, and no imaginary roots different from them, does it have two real roots?

Understand how to solve a quartic equation with roots 2 ± 3i and determine if it has two real roots. This problem delves into complex numbers, quartic equations, and the fundamental theorem of algebra. Suitable for high school students in Grades 10-12.

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