To find the zeroes of the function \( y = x^4 + 29x^2 + 100 \), we solve the equation:

\[
x^4 + 29x^2 + 100 = 0
\]

Let's perform a substitution to simplify the equation. Set \( z = x^2 \). The equation becomes:

\[
z^2 + 29z + 100 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

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

Where \( a = 1 \), \( b = 29 \), and \( c = 100 \). Plugging in these values:

\[
z = \frac{-29 \pm \sqrt{29^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1}
\]
\[
z = \frac{-29 \pm \sqrt{841 - 400}}{2}
\]
\[
z = \frac{-29 \pm \sqrt{441}}{2}
\]
\[
z = \frac{-29 \pm 21}{2}
\]

This gives us two values for \( z \):

\[
z_1 = \frac{-29 + 21}{2} = \frac{-8}{2} = -4
\]
\[
z_2 = \frac{-29 - 21}{2} = \frac{-50}{2} = -25
\]

Since \( z = x^2 \), we have \( x^2 = -4 \) and \( x^2 = -25 \).

However, these equations have no real solutions because the square of a real number cannot be negative. Thus, the solutions are complex:

For \( x^2 = -4 \):
\[
x = \pm 2i
\]

For \( x^2 = -25 \):
\[
x = \pm 5i
\]

So, the zeroes of the function are:

\[
x = 2i, -2i, 5i, -5i
\]

Would you like more details on this process or have any questions?

Here are 5 related questions for further exploration:
1. What are the zeroes of the function \( y = x^4 + 9x^2 + 20 \)?
2. How do we solve a quadratic equation using the quadratic formula?
3. What is the significance of complex roots in a polynomial?
4. Can a polynomial of degree 4 have only real zeroes? Provide an example.
5. How do we factor polynomials with complex roots?

**Tip:** When dealing with higher degree polynomials, substituting variables can simplify the problem into a more manageable quadratic equation.

Explore how to find the zeroes of the polynomial equation x^4 + 29x^2 + 100. Learn about substitution, quadratic equations, and complex number solutions. Suitable for advanced high school students and college-level math enthusiasts.

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