Solve sqrt(4 + 2x - x^2) = x - 2.

Step-by-step solution

1. 1

   Write down the sign condition before squaring. A real square root is never negative, so whatever $x$ turns out to be, the right side must satisfy

   $x-2\ge 0\quad\Longrightarrow\quad x\ge 2$

   Recording this now is what lets you throw out false roots later instead of being fooled by them.

2. 2

   Square both sides. Squaring is not reversible, so it can create solutions that do not solve the original equation — that is precisely why step 1 exists:

   $4+2x-x^{2}=(x-2)^{2}=x^{2}-4x+4$

3. 3

   Collect everything on one side and factor.

   $4+2x-x^{2}-x^{2}+4x-4=0\quad\Longrightarrow\quad 6x-2x^{2}=0$

   $2x(3-x)=0\quad\Longrightarrow\quad x=0\ \text{or}\ x=3$

4. 4

   Filter the candidates through the condition $x\ge2$. The value $x=0$ fails it immediately. Substituting anyway shows why: the left side is $\sqrt{4}=2$ while the right side is $0-2=-2$, so $2=-2$ is false. It is an artifact of squaring, not a solution.

5. 5

   Verify the surviving candidate in the original equation. At $x=3$ the radicand is $4+6-9=1$, so the left side is $\sqrt{1}=1$; the right side is $3-2=1$. Both sides equal $1$, so $x=3$ is genuine.

6. 6

   State the solution.

   $x=3$