We are given the equation:

$x^4 + x^2 - 1 = 0$

To solve this, we can make a substitution to simplify the equation. Let $y = x^2$, then the equation becomes:

$y^2 + y - 1 = 0$

This is a quadratic equation, and we can solve it using the quadratic formula:

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

For the equation $y^2 + y - 1 = 0$, we have $a = 1$, $b = 1$, and $c = -1$. Substituting these values into the quadratic formula:

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

$y = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$

Thus, the two possible values for $y$ are:

$y = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad y = \frac{-1 - \sqrt{5}}{2}$

Since $y = x^2$, and $y \geq 0$ (because $x^2 \geq 0$ for all real $x$), we discard the negative root $y = \frac{-1 - \sqrt{5}}{2}$, as it is negative. Thus, we only keep:

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

Now, solve for $x$. Since $y = x^2$:

$x^2 = \frac{-1 + \sqrt{5}}{2}$

Taking the square root of both sides:

$x = \pm \sqrt{\frac{-1 + \sqrt{5}}{2}}$

Thus, the real solutions to the equation $x^4 + x^2 - 1 = 0$ are:

$x = \pm \sqrt{\frac{-1 + \sqrt{5}}{2}}$

Would you like any further details or clarification?

Here are 5 related questions:

1. How do you solve higher-degree polynomial equations?
2. Can complex numbers arise in quadratic equations like this one?
3. What is the significance of the quadratic formula's discriminant?
4. How does factoring compare with the quadratic formula for solving equations?
5. What types of functions are involved in fourth-degree polynomial equations?

Tip: Always check for extraneous solutions when solving equations that involve squaring both sides.