The given equation is:

$x^4 + 3x^2 - 4 = 0$

Step 1: Substituting $y = x^2$

Let $y = x^2$, so that $x^4 = y^2$. The equation then transforms into a quadratic equation:

$y^2 + 3y - 4 = 0$

Step 2: Solving the Quadratic Equation

We solve for $y$ using factorization:

$y^2 + 3y - 4 = (y + 4)(y - 1) = 0$

Setting each factor to zero:

$y + 4 = 0 \quad \Rightarrow \quad y = -4$ $y - 1 = 0 \quad \Rightarrow \quad y = 1$

Since $y = x^2$, and $x^2$ cannot be negative, we discard $y = -4$ and keep $y = 1$.

Step 3: Solving for $x$

$x^2 = 1$

Taking the square root:

$x = \pm 1$

Final Answer:

$x = \pm 1$

Would you like a more detailed explanation? Here are some related questions:

1. How do we determine when substitution is useful in solving equations?
2. What happens if the quadratic equation had no real solutions?
3. How can we solve higher-degree polynomials without substitution?
4. What if the equation had a different constant term?
5. How do we check our solutions in the original equation?

Tip: Always check for extraneous solutions when solving equations involving even powers.