The equation provided is:

\[
x^4 - 10x^2 + 9 = 0
\]

This is a quadratic equation in terms of \( x^2 \). Let's solve it step by step:

### Step 1: Substitute \( y = x^2 \)
Let \( y = x^2 \). The equation then becomes:

\[
y^2 - 10y + 9 = 0
\]

### Step 2: Solve the quadratic equation for \( y \)
We can solve this quadratic equation using the quadratic formula:

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

For the equation \( y^2 - 10y + 9 = 0 \), \( a = 1 \), \( b = -10 \), and \( c = 9 \). Plugging in these values:

\[
y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(9)}}{2(1)}
\]

\[
y = \frac{10 \pm \sqrt{100 - 36}}{2}
\]

\[
y = \frac{10 \pm \sqrt{64}}{2}
\]

\[
y = \frac{10 \pm 8}{2}
\]

This gives us two solutions for \( y \):

\[
y = \frac{10 + 8}{2} = \frac{18}{2} = 9
\]
\[
y = \frac{10 - 8}{2} = \frac{2}{2} = 1
\]

### Step 3: Back-substitute \( y = x^2 \)
Now, substitute back \( y = x^2 \) to solve for \( x \):

For \( y = 9 \):
\[
x^2 = 9
\]
\[
x = \pm 3
\]

For \( y = 1 \):
\[
x^2 = 1
\]
\[
x = \pm 1
\]

### Final Solution:
The solutions for \( x \) are:

\[
x = \pm 3 \quad \text{and} \quad x = \pm 1
\]

So, the final answer is \( \boxed{x = \pm 3, \pm 1} \).

Would you like any additional details or have any questions?

Here are some related questions:

1. How can you verify the solutions \( x = \pm 3, \pm 1 \) in the original equation?
2. What if the original equation had complex roots? How would that change the solution?
3. How does substitution simplify solving higher degree polynomial equations?
4. What are the conditions under which a quadratic in terms of \( x^2 \) has real solutions?
5. How would the solution change if the equation had a term like \( x^3 \) instead of \( x^2 \)?

**Tip:** Substitution is a powerful technique when dealing with polynomial equations. It reduces the degree of the equation, making it easier to solve.

Learn how to solve the quadratic equation x^4 - 10x^2 + 9 = 0 with a detailed step-by-step solution. This problem involves using the substitution method to transform the equation into a simpler quadratic form. Suitable for advanced high school students.

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