Question

Solve the equation. Determine the solution(s) and any extraneous solution.
√x=x-2

Answer 1

To solve the equation \( \sqrt{x} = x - 2 \), we will first isolate the square root and then square both sides to eliminate the square root.

1. **Isolate the square root and square both sides:**
\[
\sqrt{x} = x - 2
\]
Square both sides:
\[
(\sqrt{x})^2 = (x - 2)^2
\]
This gives us:
\[
x = (x - 2)(x - 2)
\]
Expanding the right-hand side:
\[
x = x^2 - 4x + 4
\]

2. **Rearranging the equation:**
Move all terms to one side:
\[
0 = x^2 - 4x + 4 - x
\]
Simplify:
\[
0 = x^2 - 5x + 4
\]
This is a quadratic equation in the standard form.

3. **Factor the quadratic:**
Rewrite it as:
\[
0 = (x - 1)(x - 4)
\]

4. **Setting each factor to zero gives the solutions:**
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

Now we have two potential solutions: \( x = 1 \) and \( x = 4 \).

5. **Check for extraneous solutions:**
We need to check each potential solution in the original equation \( \sqrt{x} = x - 2 \).

- For \( x = 1 \):
\[
\sqrt{1} = 1 - 2 \quad \Rightarrow \quad 1 = -1 \quad \text{(false)}
\]
So, \( x = 1 \) is an extraneous solution.

- For \( x = 4 \):
\[
\sqrt{4} = 4 - 2 \quad \Rightarrow \quad 2 = 2 \quad \text{(true)}
\]
So, \( x = 4 \) is a valid solution.

6. **Conclusion:**
The valid solution to the equation \( \sqrt{x} = x - 2 \) is \( \boxed{4} \). The solution \( x = 1 \) is an extraneous solution.