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

The given equation is:

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

### Substitution:
Let \( u = (x + 1)^{\frac{1}{2}} \). This implies that:
- \( u^2 = x + 1 \)
- \( (x + 1)^{\frac{3}{2}} = u^3 \)
- \( (x + 1)^{\frac{5}{2}} = u^5 \)

Substituting these expressions into the equation:

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

### Factoring:
This is now a polynomial equation in \( u \):
\[
u(4 - 5u^2 + u^4) = 0
\]

The solutions are:
1. \( u = 0 \), or
2. \( 4 - 5u^2 + u^4 = 0 \)

#### Case 1: \( u = 0 \)
If \( u = 0 \), then:
\[
(x + 1)^{\frac{1}{2}} = 0 \quad \Rightarrow \quad x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

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

This factors as:
\[
(v - 4)(v - 1) = 0
\]

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

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

### Back-substitution to find \( x \):
Since \( u = (x + 1)^{\frac{1}{2}} \), we can solve for \( x \) in each case:

- If \( u = 2 \), then \( (x + 1)^{\frac{1}{2}} = 2 \), so \( x + 1 = 4 \), which gives \( x = 3 \).
- If \( u = -2 \), there is no solution because \( (x + 1)^{\frac{1}{2}} \) cannot be negative.
- If \( u = 1 \), then \( (x + 1)^{\frac{1}{2}} = 1 \), so \( x + 1 = 1 \), which gives \( x = 0 \).
- If \( u = -1 \), there is no solution for the same reason as before.

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

Would you like further details on any step?

### 5 Related Questions:
1. How do you solve higher-degree polynomial equations?
2. What are common substitution techniques in algebra?
3. How can factoring be used to simplify equations?
4. How does the quadratic formula work for non-standard quadratic equations?
5. What is the relationship between roots of polynomials and their factored form?

### Tip:
When solving equations involving square roots, always check for extraneous solutions that arise from squaring both sides.

This solution demonstrates how to solve the equation 4(x + 1)^(1/2) - 5(x + 1)^(3/2) + (x + 1)^(5/2) = 0 using substitution and factoring techniques. It includes key algebraic concepts and detailed steps for students in Grades 10-12.

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