To solve the equation \( x^{4/3} - x^{2/3} - 5 = 0 \), we can make a substitution to simplify it. Let's set:

\[
y = x^{2/3}
\]

Then \( x^{4/3} \) becomes \( y^2 \). Substituting into the original equation gives:

\[
y^2 - y - 5 = 0
\]

This is a quadratic equation in terms of \( y \). We can solve it using the quadratic formula:

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

where \( a = 1 \), \( b = -1 \), and \( c = -5 \). Plugging in these values:

\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-5)}}{2(1)} = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}
\]

So, the two solutions for \( y \) are:

\[
y = \frac{1 + \sqrt{21}}{2} \quad \text{and} \quad y = \frac{1 - \sqrt{21}}{2}
\]

Since \( y = x^{2/3} \) and \( x^{2/3} \geq 0 \) for real numbers \( x \), we discard the negative root \( \frac{1 - \sqrt{21}}{2} \) as it results in a negative value.

Therefore, the valid solution for \( y \) is:

\[
y = \frac{1 + \sqrt{21}}{2}
\]

Now, substitute back \( y = x^{2/3} \):

\[
x^{2/3} = \frac{1 + \sqrt{21}}{2}
\]

To solve for \( x \), raise both sides to the power of \( \frac{3}{2} \):

\[
x = \left( \frac{1 + \sqrt{21}}{2} \right)^{3/2}
\]

This is the real solution to the equation. If you need the numerical value or further details, let me know!

Would you like a more detailed explanation, or do you have any questions?

### Relative Questions:
1. How do we derive the quadratic formula used for solving quadratic equations?
2. What are the properties of exponents when solving equations like this?
3. Can we find the complex roots for this equation?
4. How can we graphically represent the function \( f(x) = x^{4/3} - x^{2/3} - 5 \)?
5. What other substitution techniques can simplify equations involving fractional exponents?

### Tip:
When dealing with equations involving fractional exponents, consider substitution to reduce the problem to a simpler algebraic form, like a quadratic equation.

Learn how to solve the equation x^(4/3) - x^(2/3) - 5 = 0 through substitution and the quadratic formula. This problem explores fractional exponents and quadratic equations, suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation