Let's solve the equation $x^4 - 3x^2 + 1 = 0$ to verify the solutions.

First, we can make a substitution to simplify the equation. Let $y = x^2$. Then the equation becomes: $y^2 - 3y + 1 = 0$

This is a quadratic equation in $y$. We can solve it using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -3$, and $c = 1$: $y = \frac{3 \pm \sqrt{9 - 4}}{2}$ $y = \frac{3 \pm \sqrt{5}}{2}$

So the solutions for $y$ are: $y_1 = \frac{3 + \sqrt{5}}{2}$ $y_2 = \frac{3 - \sqrt{5}}{2}$

Recall that $y = x^2$, so we need to find the square roots of these solutions to get $x$: $x = \pm \sqrt{y_1} = \pm \sqrt{\frac{3 + \sqrt{5}}{2}}$ $x = \pm \sqrt{y_2} = \pm \sqrt{\frac{3 - \sqrt{5}}{2}}$

These are the four real solutions to the equation $x^4 - 3x^2 + 1 = 0$.

To compare with your solutions: $\sqrt{\frac{3 + \sqrt{5}}{2}} \approx 1.618$ $\sqrt{\frac{3 - \sqrt{5}}{2}} \approx 0.618$

Your solutions, $\pm \sqrt{1.3} \approx \pm 1.14$ and $\pm \sqrt{0.175} \approx \pm 0.42$, are not accurate. The correct solutions are:

$\pm \sqrt{\frac{3 + \sqrt{5}}{2}} \approx \pm 1.618$ $\pm \sqrt{\frac{3 - \sqrt{5}}{2}} \approx \pm 0.618$

Therefore, the correct real solutions are: $\pm \sqrt{\frac{3 + \sqrt{5}}{2}}$ $\pm \sqrt{\frac{3 - \sqrt{5}}{2}}$

Would you like more details on this solution or have any questions?

Here are 8 related questions to further explore the topic:

1. How can we determine the roots of a polynomial equation in general?
2. What is the significance of the quadratic formula in solving polynomial equations?
3. Can all polynomial equations be solved by factoring?
4. What are complex roots, and when do they occur?
5. How do you verify if the obtained solutions are correct?
6. What is the relationship between the degree of a polynomial and the number of its roots?
7. Can you explain the process of completing the square for solving quadratic equations?
8. How does Descartes' Rule of Signs help in determining the number of real roots of a polynomial?

Tip: When solving higher-degree polynomials, consider using substitution to reduce the equation to a simpler form, such as a quadratic equation. This can make finding the solutions more manageable.