If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?

Question

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 $x$?

Answer 1

$$(x-a)^2 = (x+a)^2$$ $$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$ $$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$ $$-2ax = 2ax$$ $$-a = a$$ Note that this statement is only true when $a=0$, which is thus your solution.

Answer 2

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 $a=0$.

Answer 3

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 $a = 0$.

Answer 4

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.

Answer 5

$$\frac{(x-a)^2}{(x+a)^2}=1$$ $$\left(\frac{x-a}{x+a}\right)^2=1$$

$$\left(1-\frac{2a}{x+a}\right)^2=1$$ it is clear that $\frac{2a}{x+a}$ should be equal to zero

so, the $$a=0$$

Answer 6

$$(x-a)^2 = (x+a)^2$$ 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) $x = 0$.

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

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

Answer 7

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$