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The equation you have is:

\[
6x^4 - 17x^2 + 5 = 0
\]

This is a quartic equation, but it can be solved as a quadratic equation in terms of \(x^2\). Let's proceed step by step.

1. Let \(y = x^2\). So the equation becomes:
\[
6y^2 - 17y + 5 = 0
\]
This is a quadratic equation in \(y\).

2. Solve this quadratic using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 6\), \(b = -17\), and \(c = 5\).

Let's compute the solutions for \(y\).The solutions for \( y \) are:

\[
y_1 = 2.5 \quad \text{and} \quad y_2 = 0.3333
\]

Now, since \( y = x^2 \), we substitute back to find \( x \).

For \( y_1 = 2.5 \):
\[
x^2 = 2.5 \quad \Rightarrow \quad x = \pm \sqrt{2.5}
\]

For \( y_2 = 0.3333 \):
\[
x^2 = 0.3333 \quad \Rightarrow \quad x = \pm \sqrt{0.3333}
\]

Let's compute these square roots to find the values of \( x \).It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can manually compute the square roots as follows:

- \( x_1 = \pm \sqrt{2.5} \approx \pm 1.5811 \)
- \( x_2 = \pm \sqrt{0.3333} \approx \pm 0.5774 \)

So the solutions for \( x \) are approximately:
\[
x = \pm 1.5811, \pm 0.5774
\]

Would you like more details or have any questions about this process?

Here are 5 related questions:
1. How do you solve other types of quartic equations?
2. What is the discriminant in a quadratic equation, and how does it determine the nature of solutions?
3. Can you solve cubic equations similarly to this method?
4. How are real and complex roots determined in quadratic and quartic equations?
5. Can a quartic equation have more than four real roots?

**Tip:** Always check the discriminant in a quadratic equation to understand whether the solutions are real or complex.

Learn how to solve the quartic equation 6x^4 - 17x^2 + 5 using the quadratic formula and substitution. This problem explores algebraic techniques, quadratic formulas, and higher-order polynomials. Suitable for high school students in Grades 10-12.

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