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left parenthesis x plus 2 right parenthesis Superscript 2 divided by 5 Baseline minus 6 left parenthesis x plus 2 right parenthesis Superscript 1 divided by 5 Baseline plus 8 equals 0


You have the following equation to solve:

\[
\frac{(x + 2)^{2/5}}{5} - 6(x + 2)^{1/5} + 8 = 0
\]

### Step 1: Substitution
Let \( u = (x + 2)^{1/5} \), so that:
\[
u^2 = (x + 2)^{2/5}
\]

This simplifies the equation to:
\[
\frac{u^2}{5} - 6u + 8 = 0
\]

### Step 2: Eliminate the fraction
Multiply the entire equation by 5 to eliminate the fraction:
\[
u^2 - 30u + 40 = 0
\]

### Step 3: Solve the quadratic equation
The quadratic equation is:
\[
u^2 - 30u + 40 = 0
\]

You can solve this using the quadratic formula:
\[
u = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(1)(40)}}{2(1)}
\]
\[
u = \frac{30 \pm \sqrt{900 - 160}}{2}
\]
\[
u = \frac{30 \pm \sqrt{740}}{2}
\]
\[
u = \frac{30 \pm \sqrt{740}}{2}
\]
Since \( \sqrt{740} \approx 27.2 \), this gives:
\[
u = \frac{30 \pm 27.2}{2}
\]

Thus, the two possible solutions for \( u \) are:
\[
u_1 = \frac{30 + 27.2}{2} \approx 28.6
\]
\[
u_2 = \frac{30 - 27.2}{2} \approx 1.4
\]

### Step 4: Solve for \( x \)
Recall that \( u = (x + 2)^{1/5} \), so:
1. For \( u_1 = 28.6 \), we have:
   \[
   (x + 2)^{1/5} = 28.6
   \]
   Raise both sides to the power of 5:
   \[
   x + 2 = 28.6^5
   \]
   \[
   x = 28.6^5 - 2
   \]
   \( 28.6^5 \) is a large number (approximately 22858199.58), so:
   \[
   x \approx 22858199.58 - 2 = 22858197.58
   \]

2. For \( u_2 = 1.4 \), we have:
   \[
   (x + 2)^{1/5} = 1.4
   \]
   Raise both sides to the power of 5:
   \[
   x + 2 = 1.4^5
   \]
   \[
   x = 1.4^5 - 2
   \]
   \( 1.4^5 \approx 5.37824 \), so:
   \[
   x \approx 5.37824 - 2 = 3.37824
   \]

### Final Solutions:
The solutions are approximately:
\[
x \approx 22858197.58 \quad \text{or} \quad x \approx 3.38
\]

Would you like more details on any of the steps or have any further questions?

### Related Questions:
1. How do you solve quadratic equations in general?
2. What is the significance of the substitution method in solving complex equations?
3. What other methods can simplify fractional exponents?
4. How does the quadratic formula work and why is it useful?
5. How do you approximate large powers of decimal numbers?

**Tip:** Always check both solutions in the original equation to verify their correctness.

(x + 2)^(2/5) / 5 - 6(x + 2)^(1/5) + 8 = 0

This detailed solution explains how to solve the equation (x + 2)^(2/5) / 5 - 6(x + 2)^(1/5) + 8 = 0 using substitution and the quadratic formula. Explore algebraic concepts such as fractional exponents and quadratic equations, ideal for students in Grades 10-12.

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