<s>[INST] 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?
 [/INST] The standard way to solve equations with square roots is to use this rule:
$$ \sqrt A=B\iff (A=B^2\quad\textbf{and}\quad B\ge 0). $$
So here you obtain
$$\sqrt{x-2}=x-4\iff x-2=x^2-8x+16\;\text{and}\; x\ge 4\iff(x-3)(x-6)=0\;\text{and}\; x\ge 4,$$
which shows there's only one root: $6$.
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