<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] By definition $\sqrt{}$ is always the non-negative root.  Every positive number has exactly two square roots equal in magnitude, on positive and one negative.
$\sqrt{25} =5$ and $\sqrt {25} \ne -5$.  But both $5$ and $-5$ are solutions to $x^2 = 25$.
To solve an equation $x^2 = k$ there will be two answers.  One is $\sqrt k$ and $\sqrt k > 0$ and the other is $-\sqrt{k}$ and $-\sqrt {k} < 0$
So if you try to solve an equation by "squaring both sides", you will be changing the equation to allow for two different square roots that were not part of the original problem.  This is called superfluous solutions.
So to solve
$\sqrt {x-2} = x- 4$
Is not just to solve $x-2 = (x-4)^2$ but is to ALSO solve $x - 4 \ge 0$.
So you did $\sqrt{x-2}^2 = (x-4)^2$.  But that adds in the negative solution as well. 
$x^2 - 8x + 16 = x-2$ and $x^2 - 9x + 18 = (x-6)(x-3)$ so both of those solve $x-2 = (x-4)^2$ but only one of them solve $\sqrt{x-2} = x-4$. (Because we must have $x-4 \ge 0$.)
$\sqrt{6-2} \overset?= 6-4$
$\sqrt{4} \overset?= 2$
$2 \overset \checkmark = 2$ check.  $6$ is an answer and $6-4 > 0$.
$\sqrt{3-2} \overset?= 3-4$
$\sqrt{1} \overset?= -1$
$1 \ne -1$. No!  $\sqrt{1} = 1$.  $\sqrt{1} \ne -1$.   And $3-2 < 0$.
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