<s>[INST] (Highschool Pre-calculus) Solving quadratic via completing the square I'm trying to solve the following equation by completing the square:
$x^2 - 6x = 16$ 
The correct answer is -6,1. This is my attempt:
$x^2 - 6x = 16$
$(x - 3)^2 = 16$
$(x - 3)^2 = 25$ 
$\sqrt(x -3)^2 = \sqrt(25)$
$x - 3 = \pm5$
$x =\pm5 - 3$
$x = -8,2$
I did everything according to what I know,but my answer was obviously wrong. Any help is appreciated. Thanks
 [/INST] I am just going to do the thing first and we will talk about what you did after.
Note that $x^2-6x\sim (x-3)^2$. In fact
$$\begin{align}(x-3)^2=x^2-6x+9&=(x^2-6x)+9\\\Rightarrow x^2-6x&=(x-3)^2-9.\end{align}$$
Therefore if
$$x^2-6x=16$$ then
$$\begin{align}(x-3)^2-9&=16
\\\Rightarrow (x-3)^2&=25
\\ \Rightarrow x-3&=\pm5
\\ \Rightarrow x&=8\text{ or }-2.
\end{align}$$
The problems with what you did are as follows:


*

*If $x^2-6x=16$ it does not follow that $(x-3)^2=16$.

*You don't need to take the square roots of both sides necessarily. If $x^2=a$ then $x=\pm \sqrt{a}$ is a perfectly good implication.

*At the end you have two equations... $x-3=5$ and $x-3=-5$. To get rid of the three, remember you want $x=\dots$ add three to both sides, e.g.
$$\begin{align}
x-3&=5
\\\Rightarrow x-3+3&=5+3
\\\Rightarrow x+0&=8
\\ \Rightarrow x&=8.
\end{align}$$


You can do this because if two quantities are equal, and you add the same thing to both of them, then they are still equal.
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