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The given equation is:

\[
x^4 - 13x^2 + 36 = 0
\]

This is a quartic equation, but it can be solved by treating it as a quadratic in terms of \(x^2\). Let's rewrite it using a substitution:

Let \( y = x^2 \), then the equation becomes:

\[
y^2 - 13y + 36 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
y = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(1)(36)}}{2(1)}
\]

\[
y = \frac{13 \pm \sqrt{169 - 144}}{2}
\]

\[
y = \frac{13 \pm \sqrt{25}}{2}
\]

\[
y = \frac{13 \pm 5}{2}
\]

Thus, we get two possible values for \( y \):

1. \( y = \frac{13 + 5}{2} = 9 \)
2. \( y = \frac{13 - 5}{2} = 4 \)

Since \( y = x^2 \), we now solve for \( x \):

For \( y = 9 \), \( x^2 = 9 \), so \( x = \pm 3 \).

For \( y = 4 \), \( x^2 = 4 \), so \( x = \pm 2 \).

Thus, the solutions for \( x \) are:

\[
x = \pm 3, \pm 2
\]

So the full set of solutions is \( x = -3, -2, 2, 3 \).

Would you like more details on any of the steps? Here are some related questions:

1. How does substitution help in solving quartic equations like this?
2. Could this method work for any quartic equation?
3. What other methods can be used to solve quartic equations?
4. How do we apply the quadratic formula correctly in complex cases?
5. How would this process change if the quadratic had no real solutions?

**Tip:** When solving higher-order polynomials, always check if you can simplify the equation by substitution or factoring. This can often make solving much easier!

Learn how to solve the quartic equation x^4 - 13x^2 + 36 = 0 using substitution and the quadratic formula. This step-by-step solution covers key algebraic concepts, including solving higher-order polynomials. Suitable for students in Grades 10-12.

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