The ± sign in square root As I was doing a SAT question when I came across this question:
$\sqrt {x-a} = x-4$
       If $a=2$,what is the solution set of the equation?
Options


*

*{$3,6$}

*{$2$}

*{$3$}

*{$6$}     Correct Answer


I evaluated the equation and got $0=(x-3)(x-6)$If you put those number in the equation, you should get:
For 3:
$\sqrt {3-2} = 3-4$
Since $\sqrt {1} = ±1$  
$±1 = -1$
For 6:
$\sqrt {6-2} = 6-4$
Since $\sqrt {4} = ±2$  
$±2 = 2$
For the answer, they(SAT) evaluated $\sqrt {1}$ as $\sqrt {1} = -1$ and $\sqrt {4}$ as $\sqrt {4} = 2$ Why is it that $\sqrt {1}$ is equal to $-1$ and not $1$ and why $\sqrt {4}$ is equal to $2$ and not $-2$ Why isn't the solution set {$3,6$} a correct answer?

You factored it correctly, but your mistake is considering both roots of $x$. When we say $\sqrt{x}$, we refer to the principal root of $x$; that is, the $positive$ square root of $x$  (issue in red, corrections in blue):
For 3:
$\sqrt {3-2} = 3-4$
$\sqrt {1} = \color{red}{±}1$ 
$\color{blue}{\sqrt {1} = -1}$
$\color{blue}{1 = -1 \, \, \, \, \, \, \, \text{false; not in solution set}}$
For 6:
$\sqrt {6-2} = 6-4$
$\sqrt {4} = \color{red}{±}2$
$\color{blue}{\sqrt {4} = 2}$
$\color{blue}{2 = 2 \, \, \, \, \, \, \, \text{true; in solution set}}$
The extraneous solution $x=3$ was brought about when you squared the radical, because squaring removes the restriction that $\sqrt{u} \geq 0$.
