The equation in the image is:

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

Step 1: Substitution to Simplify

We can make a substitution to simplify the equation. Let $y = x^2$. This transforms the equation into a quadratic equation:

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

Step 2: Solving the Quadratic Equation

Now, we solve the quadratic equation $y^2 + 3y - 4 = 0$ using the quadratic formula:

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

Here, $a = 1$, $b = 3$, and $c = -4$. Plugging in these values:

$y = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)}$

$y = \frac{-3 \pm \sqrt{9 + 16}}{2}$

$y = \frac{-3 \pm \sqrt{25}}{2}$

$y = \frac{-3 \pm 5}{2}$

This gives us two possible values for $y$:

$y = \frac{-3 + 5}{2} = 1$

$y = \frac{-3 - 5}{2} = -4$

Step 3: Back-Substitution and Solving for $x$

Recall that $y = x^2$. Now, we substitute back to find $x$.

1. For $y = 1$:

$x^2 = 1 \implies x = \pm 1$

2. For $y = -4$:

$x^2 = -4$

Since $x^2 = -4$ has no real solutions (as the square of a real number cannot be negative), this does not contribute any real roots.

Final Answer

The real solutions are $x = 1$ and $x = -1$. There are no extraneous solutions in this case because all steps were algebraically valid.

Would you like more details or have any further questions?

Related Questions:

1. How do we handle quadratic equations with complex roots?
2. What is the significance of extraneous solutions in algebraic equations?
3. How can substitution simplify higher-degree polynomial equations?
4. What are the conditions under which a quadratic equation has no real solutions?
5. How would the solution differ if the quadratic was non-factorable by standard means?

Tip:

When solving polynomial equations, always consider possible substitutions that reduce the equation to a more familiar form, such as a quadratic equation.