The correct way of solving for $a$ in $a^2 = b^2$ Given an equation $a^2=b^2$, solving for $a$ would give
$$\sqrt {a^2} = \sqrt {b^2}$$
$$|a| = |b|$$
$$a = \pm |b|$$
Is there a way for me to solve for $a$ without having the absolute value sign on $b$? I believe the answer should be $a = \pm b$ and this is my goal of doing all this. If I divide $a = \pm |b|$ into cases and try to solve for b:
Case one: $a = |b|$
$b = a$ or $b = -a$. Put it another way, $b = \pm a$
Case two: $a = -|b|$
Solving for $|b|$ would give us $|b| = -a$. Hence,
$b = (-a)$ or $b = -(-a)$. Put this another way, $b = \pm (-a)$

So, now I have $b = \pm a$ or $b = \pm (-a)$. I believe if I try to solve for $a$ on both equations, I'll get (correct me if I'm wrong):
$$a = \pm b$$
or
$$a = \pm (-b)$$
So, is it correct if I say that solving for $a$ in $a^2 = b^2$ gives me $a = \pm b$? Even though it actually gives me $a = \pm b$ or $a = \pm (-b)$. If yes, how so since there are two possibilities. I'm also aware that $\pm b \neq \pm (-b)$ which makes me even more confuse about it. I appreciate any help.
 A: $$a^2=b^2\iff a^2-b^2=(a-b)(a+b)=0.$$
So $$a-b=0\text{ or }a+b=0.$$
A: Further to user1015917's answer, I'd like to point out that even though my instinct is to continue from your first chunk of working by simply writing \begin{gather}a = &\pm|b|\\=&{\pm}(\pm b)\tag#\\=&{\pm}b,\end{gather} there is a caveat.
The intermediate expression $$\pm (\pm b)$$ is arguably ambiguous as to whether the top and bottom signs are meant to separately/independently correspond, i.e., is the expression meant to be understood as \begin{align}&{\pm}(\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&b,\end{align} or whether \begin{align}&{\pm} (\pm b)\\=&{+}(+b)\:\:\text{or}\:\:{+}({-}b)\:\:\text{or}\:\:{-}(+b)\:\:\text{or}\:\:{-}({-}b)\\=&{\pm}b.\end{align}
In $(\#)$ and the final line of this answer, I mean the second interpretation, however, typically, the expression $$(\pm \,a\mp b)$$ is interpreted the first way, as $$\pm(a-b).$$
Summarising: an expression containing multiple occurrences of $\pm$ and/or $\mp$ may be disambiguated from its context.
