Question

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

Explanation:

Step1: Test \( x = 2 \)

Substitute \( x = 2 \) into the equation \( x = 2 + \sqrt{x - 2} \). The square root term becomes \( \sqrt{2 - 2} = \sqrt{0} = 0 \). So the right - hand side is \( 2+0 = 2 \), and the left - hand side is also \( 2 \). But we need to check the domain of the square root function. The expression under the square root \( x - 2\geq0\), so \( x\geq2 \). However, let's check the next value too.

Step2: Test \( x = 3 \)

Substitute \( x = 3 \) into the equation. The square root term is \( \sqrt{3 - 2}=\sqrt{1} = 1 \). The right - hand side is \( 2 + 1=3 \), and the left - hand side is also \( 3 \). Now, we need to verify if both solutions are valid. Wait, when \( x = 2 \), the original equation is \( 2=2+\sqrt{2 - 2}\), which is \( 2 = 2+0\), but if we solve the equation algebraically:
Let \( y=\sqrt{x - 2}\), then the equation becomes \( x=2 + y \), and since \( y=\sqrt{x - 2}\), we have \( x-2=y^{2}\). Substitute \( y=x - 2 \) into \( x-2=y^{2}\), we get \( x - 2=(x - 2)^{2}\). Let \( t=x - 2 \), then \( t=t^{2}\), \( t^{2}-t = 0\), \( t(t - 1)=0\), so \( t = 0\) or \( t = 1\).
If \( t = 0\), then \( x-2=0\), \( x = 2\); if \( t = 1\), then \( x-2=1\), \( x=3\). But we need to check the original equation for \( x = 2\): \( 2=2+\sqrt{2 - 2}\), \( 2=2 + 0\), which is true. For \( x = 3\): \( 3=2+\sqrt{3 - 2}\), \( 3=2 + 1\), which is also true. Wait, but the options include \( x = 2\), \( x = 3\), \( x = 2\) or \( x = 3\), and no solution. But let's check the algebraic solution again.
Wait, when we set \( y=\sqrt{x - 2}\), \( y\geq0\). When \( x = 2\), \( y = 0\) (which is non - negative). When \( x=3\), \( y = 1\) (which is non - negative). But let's go back to the substitution. Wait, maybe there is a mistake in the initial thought. Wait, if we solve \( x-2=(x - 2)^{2}\), \( (x - 2)^{2}-(x - 2)=0\), \( (x - 2)(x - 2 - 1)=0\), \( (x - 2)(x - 3)=0\), so \( x = 2\) or \( x = 3\). But let's check the original equation for \( x = 2\): \( 2=2+\sqrt{2 - 2}\), \( 2=2+0\), true. For \( x = 3\): \( 3=2+\sqrt{3 - 2}\), \( 3=2 + 1\), true. But the options are given as single options? Wait, maybe I made a mistake. Wait, the problem is written as \( x=2+\sqrt{x - 2}\). Let's check \( x = 2\): LHS = 2, RHS=2+\sqrt{0}=2, so it works. \( x = 3\): LHS = 3, RHS=2+\sqrt{1}=3, so it works. But the option "x = 2 or x = 3" is available. Wait, but maybe the question has a typo or I misread. Wait, no, let's check the options again. The options are \( x = 2\), \( x = 3\), \( x = 2\) or \( x = 3\), no solution.

Wait, but when we solve the equation \( x=2+\sqrt{x - 2}\) algebraically:

1. Isolate the square root: \( \sqrt{x - 2}=x - 2\)
2. Let \( u=\sqrt{x - 2}\), then \( u=x - 2\) and \( u\geq0\). Also, from \( u=\sqrt{x - 2}\), we have \( u^{2}=x - 2\). Substitute \( x=u + 2\) into \( u^{2}=x - 2\), we get \( u^{2}=u+2 - 2=u\), so \( u^{2}-u = 0\), \( u(u - 1)=0\), so \( u = 0\) or \( u = 1\)
- If \( u = 0\), then \( \sqrt{x - 2}=0\), so \( x-2 = 0\), \( x = 2\)
- If \( u = 1\), then \( \sqrt{x - 2}=1\), so \( x-2 = 1\), \( x = 3\)

Both \( x = 2\) and \( x = 3\) satisfy the original equation. But let's check the options. The option "x = 2 or x = 3" is present. But wait, maybe the problem was supposed to have a different form. Wait, no, according to the substitution, both values work. But let's check the arithmetic again for \( x = 2\): \( 2=2+\sqrt{2 - 2}\), \( 2=2+0\), correct. For \( x = 3\): \( 3=2+\sqrt{3 - 2}\), \( 3=2 + 1\), correct. So the solution is \( x = 2\) or \( x = 3\). But wait, maybe I made a mistake. Wait, the original equation is \( x=2+\sqrt{x - 2}\). Let's check the domain: \( x-2\geq0\Rightarrow x\geq2\). Both \( x = 2\) and \( x = 3\) are in the domain. So the solution is \( x = 2\) or \( x = 3\). But let's check the options again. The options are:

- \( x = 2\)
- \( x = 3\)
- \( x = 2\) or \( x = 3\)
- no solution

So the correct option is "x = 2 or x = 3". But wait, maybe there is an error in my calculation. Wait, no, when \( x = 2\), the equation holds, and when \( x = 3\), the equation holds. So the solution is \( x = 2\) or \( x = 3\).

Wait, but let's solve it again:

Start with \( x=2+\sqrt{x - 2}\)

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

Square both sides: \( (x - 2)^{2}=x - 2\)

Expand the left - hand side: \( x^{2}-4x + 4=x - 2\)

Bring all terms to one side: \( x^{2}-5x + 6 = 0\)

Factor the quadratic: \( (x - 2)(x - 3)=0\)

So \( x = 2\) or \( x = 3\). Now, we need to check for extraneous solutions (solutions that come from squaring both sides but do not satisfy the original equation).

For \( x = 2\):

Left - hand side: \( x = 2\)

Right - hand side: \( 2+\sqrt{2 - 2}=2+0 = 2\). So \( x = 2\) is a valid solution.

For \( x = 3\):

Left - hand side: \( x = 3\)

Right - hand side: \( 2+\sqrt{3 - 2}=2 + 1=3\). So \( x = 3\) is a valid solution.

So both \( x = 2\) and \( x = 3\) are solutions.

Answer:

\( x = 2 \) or \( x = 3 \) (the option corresponding to "x = 2 or x = 3")