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?
 A: 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: $6$.
A: The square root sign over a real number always denotes just the nonnegative square root. There's no $\pm$. If you square an equation with a square root in it you run the risk of allowing an extraneous solution.
In this example, $\sqrt{3-2} = 1 \ne 3-4$ .
There are many questions on this site with this answer, It was just easier for me to write it than to find them.
A: 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 $3-2 < 0$.
A: 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 $\sqrt{u} \geq 0$.
