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The equation given is:

\[
\sqrt{x - 10} + \sqrt{x + 5} = 5
\]

Let's solve this step-by-step.

### Step 1: Isolate one square root
Move one of the square roots to the other side of the equation. Let's isolate \(\sqrt{x + 5}\):

\[
\sqrt{x - 10} = 5 - \sqrt{x + 5}
\]

### Step 2: Square both sides
Now square both sides of the equation to eliminate the square root on the left-hand side:

\[
(\sqrt{x - 10})^2 = (5 - \sqrt{x + 5})^2
\]

This simplifies to:

\[
x - 10 = (5)^2 - 2 \cdot 5 \cdot \sqrt{x + 5} + (\sqrt{x + 5})^2
\]

Simplifying the right-hand side:

\[
x - 10 = 25 - 10\sqrt{x + 5} + (x + 5)
\]

Now, simplify the equation:

\[
x - 10 = 25 + x + 5 - 10\sqrt{x + 5}
\]

\[
x - 10 = x + 30 - 10\sqrt{x + 5}
\]

### Step 3: Simplify further
Cancel out the \(x\) terms on both sides:

\[
-10 = 30 - 10\sqrt{x + 5}
\]

Move the constant 30 to the other side:

\[
-40 = -10\sqrt{x + 5}
\]

### Step 4: Solve for the square root term
Divide both sides by \(-10\):

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

### Step 5: Square both sides again
Square both sides to remove the square root:

\[
16 = x + 5
\]

### Step 6: Solve for \(x\)
Subtract 5 from both sides:

\[
x = 11
\]

### Step 7: Check for extraneous solutions
Substitute \(x = 11\) into the original equation to check if it works:

\[
\sqrt{11 - 10} + \sqrt{11 + 5} = \sqrt{1} + \sqrt{16} = 1 + 4 = 5
\]

Since the original equation is satisfied, \(x = 11\) is a valid solution.

### Final Answer:
The solution is \(x = 11\).

---

Would you like more details on this solution, or do you have any other questions?

Here are 5 follow-up questions for further practice:
1. What would the solution be if the equation were \(\sqrt{x - 2} + \sqrt{x + 3} = 6\)?
2. Can you solve the equation \(\sqrt{x + 7} + \sqrt{x - 1} = 4\)?
3. How would the solution change if the original equation had a negative number on the right-hand side?
4. What happens if we square an equation with a square root and there are multiple square roots?
5. Why is it important to check for extraneous solutions after solving radical equations?

**Tip**: Always check for extraneous solutions when dealing with square roots, as squaring both sides of an equation can sometimes introduce solutions that don't satisfy the original equation.

Solve the equation √(x - 10) + √(x + 5) = 5.

Learn how to solve the radical equation √(x - 10) + √(x + 5) = 5. This problem involves key algebraic concepts like radical equations and square roots. Suitable for students in Grades 10-12, with a step-by-step solution provided.

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