<s>[INST] If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$? If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?
At the end when you get $4ax=0$, can I divide by $4x$ to cancel out $4$ and $x$?
 [/INST] $$(x-a)^2 = (x+a)^2$$
Expand squares:
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
Subtract $x^2+a^2$ from both sides:
$$-2ax = 2ax$$
add $2ax$ and divide by $4$:
$$0 = ax$$
So either 1) $a = 0$ or 2) $x = 0$.


*

*$$(x-0)^2 = (x+0)^2 \Leftrightarrow x^2 = x^2$$
Which is true for all x.

*$$(0-a)^2 = (0+a)^2 \Leftrightarrow (-a)^2 = (a)^2$$
Which is true for all $a$.


So either $a=0$ and $x$ can be anything or $x=0$ and $a$ can be anything.
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