To solve the equation \(\sqrt{x-3} = x-5\) using substitution, let's follow these steps:

### Step 1: Make a substitution.
Let \( u = \sqrt{x-3} \). This means:
\[
u = \sqrt{x-3} \quad \text{so} \quad u^2 = x - 3 \quad \text{or} \quad x = u^2 + 3
\]

### Step 2: Substitute \(x = u^2 + 3\) into the original equation.
Now, substitute this expression for \(x\) into the original equation:
\[
u = (u^2 + 3) - 5
\]
Simplifying the right-hand side:
\[
u = u^2 - 2
\]

### Step 3: Rearrange the equation.
Rearrange the equation to get a standard quadratic form:
\[
u^2 - u - 2 = 0
\]

### Step 4: Solve the quadratic equation.
Factor the quadratic equation:
\[
(u - 2)(u + 1) = 0
\]
So, the solutions are:
\[
u = 2 \quad \text{or} \quad u = -1
\]

### Step 5: Reverse the substitution.
Since \( u = \sqrt{x-3} \), and the square root function only gives non-negative results, we must discard \( u = -1 \), as it's not valid.

Thus, \( u = 2 \). Now substitute back:
\[
\sqrt{x - 3} = 2
\]

### Step 6: Solve for \(x\).
Square both sides:
\[
x - 3 = 4
\]
Thus, \(x = 7\).

### Step 7: Check the solution.
Substitute \(x = 7\) back into the original equation to verify:
\[
\sqrt{7 - 3} = 7 - 5
\]
\[
\sqrt{4} = 2
\]
This is true, so \(x = 7\) is the correct solution.

### Final Answer:
The solution to \(\sqrt{x-3} = x-5\) is \(x = 7\).

Would you like more details on any step? 

Here are some related questions to explore:
1. What are some other methods to solve radical equations?
2. How can we check for extraneous solutions in radical equations?
3. What happens if we use different substitutions for the same problem?
4. Can you apply substitution methods in solving quadratic equations?
5. How do extraneous solutions arise when squaring both sides of an equation?

**Tip:** Always check your solutions when solving equations involving square roots, as squaring both sides can introduce extraneous solutions.

Discover how to solve the radical equation √(x-3) = x-5 using substitution. This detailed step-by-step guide covers algebraic concepts, substitution methods, and solving quadratic equations, suitable for Grades 9-11 students.

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