smc

$x+\sqrt{x-2}=4$ 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. $6+\sqrt{6-2}=6+\sqrt{4}=6+2=8\ne 4$ $3+\sqrt{3-2}=3+\sqrt{1}=3+1=4✓$ There is $\boxed{1\text{ real root}}$