https://file.aiquickdraw.com/mathbot/file/1725865134008mwf35ho2.jpg

The equation provided in the image is:

\[
x^4 + 3x^2 - 4 = 0
\]

This is a quartic equation (degree 4) that can be solved by using a substitution method.

### Step-by-step solution:

1. **Substitution:**
   Let \( y = x^2 \). Then, the equation becomes:
   \[
   y^2 + 3y - 4 = 0
   \]

2. **Solve the quadratic equation:**
   Solve the quadratic equation \( y^2 + 3y - 4 = 0 \) using the quadratic formula:
   \[
   y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1 \), \( b = 3 \), and \( c = -4 \).

   Substituting the values:
   \[
   y = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)}
   \]
   \[
   y = \frac{-3 \pm \sqrt{9 + 16}}{2}
   \]
   \[
   y = \frac{-3 \pm \sqrt{25}}{2}
   \]
   \[
   y = \frac{-3 \pm 5}{2}
   \]

   So, we have two possible values for \( y \):
   \[
   y = \frac{-3 + 5}{2} = 1 \quad \text{or} \quad y = \frac{-3 - 5}{2} = -4
   \]

3. **Back-substitute \( y = x^2 \):**
   - If \( y = 1 \), then \( x^2 = 1 \), so \( x = \pm 1 \).
   - If \( y = -4 \), then \( x^2 = -4 \). Since there is no real solution for \( x^2 = -4 \), we discard this case in the real number system.

### Final solution:
The real solutions to the equation are:
\[
x = 1 \quad \text{or} \quad x = -1
\]

Would you like more details or further clarifications?

### Related questions:
1. How would the solution change if we allowed complex numbers?
2. What is the general method for solving quartic equations?
3. Can we graphically verify the solutions of this equation?
4. How can we use substitution in other polynomial equations?
5. What are the different ways to factorize quartic equations?

### Tip:
Always check if substitution can simplify the process of solving higher-degree equations, especially when the equation involves even powers of \(x\).

Learn how to solve the quartic equation x^4 + 3x^2 - 4 = 0 using the substitution method. This step-by-step solution covers key algebraic concepts, including the quadratic formula and substitution techniques, and is suitable for students in Grades 9-12.

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