The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$
Solution 1
$x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{\textbf{(E)} \text{1 real root}}$

Solution 2
It's not hard to note that $x=3$ simply works, as $3 + \sqrt{1} = 4$. But, $x$ is increasing, and $\sqrt{x-2}$ is increasing, so $3$ is the only root. If $x  3$, then $x + \sqrt{x-2} > 4$. Imaginary roots don't have to be considered because squaring the equation gives a quadratic, and quadratics either have only imaginary or only real roots.

Solution 3
We can create symmetry in the equation:
\[x+\sqrt{x-2} = 4\]
\[x-2+\sqrt{x-2} = 2.\]
Let $y = \sqrt{x-2}$, then we have
\[y^2+y-2 = 0\]
\[(y+2)(y-1) = 0\]
The two roots are $\sqrt{x-2} = -2, 1$.
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{\textbf{(E)} \text{1 real root}}$.
~Vndom