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 $x$?
 A: Since we know that $$(x-a)^2 = (x+a)^2$$ holds for all $x$,
we particularly know that it holds for $x=a$ as well. Plugging that in the relation yields $$0=4a^2$$ from which we conclude that $a=0$.
A: $$(x-a)^2 = (x+a)^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$-2ax = 2ax$$
$$-a = a$$
Note that this statement is only true when $a=0$, which is thus your solution.
A: Simplifying, after expansion of squares you get
$$ a \cdot x = 0 $$
either or both of them can be zero.
However you are specifically given that for all values of $x$... So do  not put a particular $ x=0. $
Only choice is $ a=0. $
Also for any function if $ f(x+a) = f(x-a)$, then $ a =0.$
Like if $\, e^ { x+a  } \sin ( x+a) = e^ { x-a  } \sin ( x-a) $, then also, $ a=0$
A: If $(x-a)^2=(x+a)^2$ for all $x$, then graphs of functions $x\mapsto (x-a)^2$ and $x\mapsto (x+a)^2$ coincide but these are just parabolas with roots at $a$ and $-a$, respectively. Since they must coincide, $a = -a$ which implies $a = 0$.
A: $$\frac{(x-a)^2}{(x+a)^2}=1$$
$$\left(\frac{x-a}{x+a}\right)^2=1$$
$$\left(1-\frac{2a}{x+a}\right)^2=1$$
it is clear that $\frac{2a}{x+a}$ should be equal to zero
so, the $$a=0$$
A: I think the other answers fit best to solve what you were thinking, but I will write this in the case you want to know more.
If we are working in a polynomial ring like $\mathbb Z _2 \left[ x \right]$, then this equality is always true, as we get that $\left(x-a \right)^2 = \left(x+a \right)^2$ for all $a\in \mathbb Z _2$. This is because in $\mathbb Z _2 = \left\{ 0,1 \right\}$ we have that any element is its own additive opposite (that is to say, $0+0=0$ and $1+1=0$), so $a=-a$ in any of the two cases.
This doesn't happen when working on $\mathbb R$, for example, but I thought it was an interesting thing to add.
A: $$(x-a)^2 = (x+a)^2$$
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.
