4 * (x + 1) ^ (1/2) - 5 * (x + 1) ^ (3/2) + (x + 1) ^ (5/2) = 0

The given equation is:

\[
4 \cdot (x + 1)^{\frac{1}{2}} - 5 \cdot (x + 1)^{\frac{3}{2}} + (x + 1)^{\frac{5}{2}} = 0
\]

Let's make a substitution to simplify the expression:

Let \( u = (x + 1)^{\frac{1}{2}} \), which means \( u^2 = x + 1 \).

Now, substitute \( u \) into the equation:

\[
4u - 5u^3 + u^5 = 0
\]

This is a polynomial equation in terms of \( u \). To solve this equation, factor it:

\[
u(4 - 5u^2 + u^4) = 0
\]

So, we have two possible cases:

### Case 1: \( u = 0 \)
If \( u = 0 \), then \( (x + 1)^{\frac{1}{2}} = 0 \), which means \( x + 1 = 0 \), so \( x = -1 \).

### Case 2: \( 4 - 5u^2 + u^4 = 0 \)
This is a quadratic equation in terms of \( u^2 \). Let \( v = u^2 \), so the equation becomes:

\[
v^2 - 5v + 4 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
v = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}
\]

\[
v = \frac{5 \pm \sqrt{25 - 16}}{2}
\]

\[
v = \frac{5 \pm \sqrt{9}}{2}
\]

\[
v = \frac{5 \pm 3}{2}
\]

Thus, \( v = 4 \) or \( v = 1 \).

Since \( v = u^2 \), we have the following possibilities for \( u \):

- If \( v = 4 \), then \( u = 2 \) or \( u = -2 \).
- If \( v = 1 \), then \( u = 1 \) or \( u = -1 \).

Now, recall that \( u = (x + 1)^{\frac{1}{2}} \), so \( u \geq 0 \). Therefore, we only consider the non-negative values of \( u \), which are \( u = 2 \) and \( u = 1 \).

### For \( u = 2 \):
\[
(x + 1)^{\frac{1}{2}} = 2 \implies x + 1 = 4 \implies x = 3
\]

### For \( u = 1 \):
\[
(x + 1)^{\frac{1}{2}} = 1 \implies x + 1 = 1 \implies x = 0
\]

### Conclusion:
The solutions to the equation are:
\[
x = -1, 0, 3
\]

Would you like more details or explanations on any part of this process?

---

Here are some follow-up questions to explore:

1. What are the general steps in solving radical equations?
2. How can we factor higher-order polynomials?
3. Can all equations with fractional exponents be solved using substitutions?
4. Why do we discard negative values for \( u \) in this context?
5. How can we verify solutions of radical equations?

**Tip:** Always check your solutions in the original equation to ensure no extraneous solutions were introduced during squaring steps!

Learn how to solve the radical equation 4(x+1)^(1/2) - 5(x+1)^(3/2) + (x+1)^(5/2) = 0 using substitution and the quadratic formula. This detailed solution explains the substitution method, polynomial factorization, and solving quadratic equations. Suitable for students in Grades 10-12.

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