Question

what is the solution of $x = 2 + \sqrt{x - 2}$?
$x = 2$
$x = 3$
no solution
$x = 2$ or $x = 3$

Explanation:

Step1: Isolate the square root

Subtract 2 from both sides: \( x - 2 = \sqrt{x - 2} \)

Step2: Let \( y = \sqrt{x - 2} \), then the equation becomes \( y^2 = y \)

Step3: Solve \( y^2 - y = 0 \)

Factor: \( y(y - 1) = 0 \), so \( y = 0 \) or \( y = 1 \)

Step4: Substitute back \( y = \sqrt{x - 2} \)

- If \( y = 0 \): \( \sqrt{x - 2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2 \)
- If \( y = 1 \): \( \sqrt{x - 2} = 1 \Rightarrow x - 2 = 1 \Rightarrow x = 3 \)

Step5: Check solutions in original equation

- For \( x = 2 \): \( 2 = 2 + \sqrt{2 - 2} \Rightarrow 2 = 2 + 0 \), which is true? Wait, no: \( 2 = 2 + 0 \) is true? Wait, \( 2 = 2 + 0 \) is \( 2 = 2 \), yes. Wait, but wait, let's check \( x = 3 \): \( 3 = 2 + \sqrt{3 - 2} \Rightarrow 3 = 2 + 1 \), which is true. Wait, but when we checked \( x = 2 \), \( 2 = 2 + \sqrt{0} \Rightarrow 2 = 2 \), which is true. But wait, let's go back. Wait, original equation: \( x = 2 + \sqrt{x - 2} \). For \( x = 2 \), left side is 2, right side is \( 2 + \sqrt{0} = 2 \), so that's true. For \( x = 3 \), left side is 3, right side is \( 2 + \sqrt{1} = 3 \), true. But wait, the options include "x=2 or x=3" and "x=2", "x=3", "no solution". Wait, but let's check again. Wait, when we isolated the square root, we had \( x - 2 = \sqrt{x - 2} \). Let \( t = \sqrt{x - 2} \), so \( t^2 = t \), so \( t(t - 1) = 0 \), so \( t = 0 \) or \( t = 1 \). So \( \sqrt{x - 2} = 0 \Rightarrow x = 2 \), \( \sqrt{x - 2} = 1 \Rightarrow x = 3 \). Both satisfy the original equation? Wait, but let's check \( x = 2 \) again: \( 2 = 2 + \sqrt{2 - 2} \Rightarrow 2 = 2 + 0 \), which is true. \( x = 3 \): \( 3 = 2 + \sqrt{1} \), true. But the option "x=2 or x=3" is there. Wait, but maybe I made a mistake. Wait, no, let's check the original equation again. Wait, the original equation is \( x = 2 + \sqrt{x - 2} \). For \( x = 2 \), LHS = 2, RHS = 2 + 0 = 2, so that's true. For \( x = 3 \), LHS = 3, RHS = 2 + 1 = 3, true. So both are solutions? But the options include "x=2 or x=3". Wait, but let's check the options. The options are:

1. \( x = 2 \)

2. \( x = 3 \)

3. no solution

4. \( x = 2 \) or \( x = 3 \)

Wait, but when I checked, both work. But maybe I made a mistake. Wait, no, let's re-express the original equation: \( x - 2 = \sqrt{x - 2} \). Let \( t = \sqrt{x - 2} \), so \( t \geq 0 \), and \( t^2 = t \), so \( t(t - 1) = 0 \), so \( t = 0 \) or \( t = 1 \). Thus, \( x = 2 + t^2 \)? Wait, no: \( t = \sqrt{x - 2} \), so \( x = t^2 + 2 \). Wait, no, \( t = \sqrt{x - 2} \Rightarrow x = t^2 + 2 \). Wait, when \( t = 0 \), \( x = 0 + 2 = 2 \). When \( t = 1 \), \( x = 1 + 2 = 3 \). So both are solutions. But let's check the original equation again with \( x = 2 \): \( 2 = 2 + \sqrt{2 - 2} \Rightarrow 2 = 2 + 0 \), which is true. With \( x = 3 \): \( 3 = 2 + \sqrt{3 - 2} \Rightarrow 3 = 2 + 1 \), true. So the solution is \( x = 2 \) or \( x = 3 \)? But wait, the options have "x=2 or x=3" as an option. Wait, but maybe I made a mistake in the check for \( x = 2 \). Wait, \( x = 2 \): left side is 2, right side is \( 2 + \sqrt{0} = 2 \), so that's correct. \( x = 3 \): left side 3, right side \( 2 + 1 = 3 \), correct. So the solution is \( x = 2 \) or \( x = 3 \). But wait, let's check the options. The options are:

- \( x = 2 \)

- \( x = 3 \)

- no solution

- \( x = 2 \) or \( x = 3 \)

So the correct answer should be \( x = 2 \) or \( x = 3 \)? Wait, but let's check again. Wait, maybe I made a mistake in the algebra. Let's start over.

Original equation: \( x = 2 + \sqrt{x - 2} \)

Subtract 2: \( x - 2 = \sqrt{x - 2} \)

Let \( u = \sqrt{x - 2} \), so \( u \geq 0 \), and the equation becomes \( u^2 = u \)

\( u^2 - u = 0 \)

\( u(u - 1) = 0 \)

So \( u = 0 \) or \( u = 1 \)

If \( u = 0 \): \( \sqrt{x - 2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2 \)

If \( u = 1 \): \( \sqrt{x - 2} = 1 \Rightarrow x - 2 = 1 \Rightarrow x = 3 \)

Check \( x = 2 \) in original equation: \( 2 = 2 + \sqrt{2 - 2} \Rightarrow 2 = 2 + 0 \), which is true.

Check \( x = 3 \) in original equation: \( 3 = 2 + \sqrt{3 - 2} \Rightarrow 3 = 2 + 1 \), which is true.

Thus, both \( x = 2 \) and \( x = 3 \) are solutions. So the answer is \( x = 2 \) or \( x = 3 \).

Wait, but the options include "x=2 or x=3" as the fourth option. So that's the correct one.

Answer:

x = 2 or x = 3 (the option labeled "x=2 or x=3")