https://file.aiquickdraw.com/mathbot/file/1727068700538egjeb2x5.jpg

The given equation is:

\[
4 + \sqrt{x - 2} = 5 \sqrt[4]{x - 2}
\]

Let's solve this step-by-step.

1. **Let \( \sqrt{x - 2} = y \):**

   This implies that \( x - 2 = y^2 \) and hence \( x = y^2 + 2 \).

2. Substitute into the equation:

   \[
   4 + y = 5 \sqrt[4]{y^2}
   \]

   Since \( \sqrt[4]{y^2} = \sqrt{|y|} \), we can write the equation as:

   \[
   4 + y = 5 \sqrt{|y|}
   \]

3. **Square both sides** to eliminate the square root:

   \[
   (4 + y)^2 = \left( 5 \sqrt{|y|} \right)^2
   \]

   \[
   16 + 8y + y^2 = 25 |y|
   \]

   Now, we have two cases to consider because of the absolute value:

   **Case 1:** \( y \geq 0 \)

   \[
   16 + 8y + y^2 = 25y
   \]

   Rearrange:

   \[
   y^2 - 17y + 16 = 0
   \]

   Factor the quadratic:

   \[
   (y - 16)(y - 1) = 0
   \]

   So, \( y = 16 \) or \( y = 1 \).

   For \( y = 16 \):

   \[
   \sqrt{x - 2} = 16 \implies x - 2 = 256 \implies x = 258
   \]

   For \( y = 1 \):

   \[
   \sqrt{x - 2} = 1 \implies x - 2 = 1 \implies x = 3
   \]

   **Case 2:** \( y < 0 \)

   In this case, squaring would give \( 25|y| = -25y \), which is not possible as it would yield a negative result. Thus, no solutions from this case.

**Final Solutions:**

1. \( x = 3 \)
2. \( x = 258 \)

Would you like further details on this problem, or do you have any questions?

---

### Follow-up Questions:
1. How do we factor quadratic equations effectively?
2. What does the absolute value represent in mathematical terms?
3. How do different values for square roots influence the solutions?
4. What techniques can be used to solve higher-order root equations?
5. What are the potential pitfalls of squaring both sides in an equation?

**Tip:** Always check your solutions in the original equation to avoid extraneous roots.

Solve the equation: 4 + √(x - 2) = 5^(4)√(x - 2)

This detailed step-by-step guide solves the equation 4 + √(x - 2) = 5^(4)√(x - 2), using properties of radicals, quadratic equations, and absolute value. Learn key algebraic concepts and the quadratic formula, suitable for high school students (Grades 9-12).

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