# 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 $x$?

• Similar question asked about an hour back Sep 17, 2015 at 4:20
• Choose $x=1$ (or $17$, or any specific number other than $0$), and you can cancel, that is, divide both sides by $4x$. Sep 17, 2015 at 4:20
• Can you write that out please. Thank you. Sep 17, 2015 at 4:24
• This equality holds all $x$. It obviously holds for $x=0$ So now, as @AndréNicolas suggested, take any $x\ne 0$ and divide both sides of your resulting equality by $x$. You should find that $a=0$. Sep 17, 2015 at 4:38
• If we set $x=0$, we cannot divide both sides by $4x$. (Actually, we need not really worry, we can do a formal division, but that's a more abstract approach). Sep 17, 2015 at 4:48

## 7 Answers

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 $a=0$.

• No it doesn't. But it should be frustrating, shouldn't it? Sep 17, 2015 at 8:06
• @mathreadler Sorry, I don't understand what you mean. Could you be a bit more precise? Sep 17, 2015 at 10:56
• Yes, $x=a$ is not true unless also $a=0$, so why do you make the assumption $x=a$? Sep 17, 2015 at 11:05
• @mathreadler : That is basic logic: We know that: For all $x\in \mathbb{R}$, the proposition $P(x)$ holds. As $a$ is a specific, fixed real number, we also have $P(a)$. Well, I set $x=a$ to get to the solution in a quick way. I will edit my answer to make it clearer. Sep 17, 2015 at 12:05
• Note that for $a=0$ the equation turns into $x^2=x^2$ which is fulfilled for any $x$. Also, there cannot be any more a's as it is a necessary requirement that the equation holds also for $x=a$. All in all, we proved that if a solution exists, it has to be $a=0$ and that indeed $a=0$ works. Sep 17, 2015 at 21:00

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

• Why the $b$? I guess it should be $x$... Sep 17, 2015 at 5:52
• @mickep Yep. I rushed through that without thinking... That's what I get for helping out on Math.SE during the middle of the night when I should be sleeping. I fixed it... nice catch. Sep 17, 2015 at 6:28
• You can't divide by $x$ if $x = 0$... A more proper solution is to rewrite to $4ax = 0$ and conclude that either a or x must be 0. If $x = 0$ any $a$ would do and if $a = 0$ any $x$ will do. Sep 17, 2015 at 8:03
• @mathreadler This way works for all $x \neq 0$. The case $x=0$ results in $a=a$, and is thus always true. Thus, accepting $a=0$ from the solution I give allows for $x=0$ to be continuous. Sep 17, 2015 at 20:52

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, $a=0$

• Great explanation. Sep 17, 2015 at 11:06

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 $a = 0$.

• if $x = 0$, $a$ could be anything so something is wrong. Sep 17, 2015 at 8:11
• @mathreadler, that's why it says "for all $x$" which basically translates to equality of functions $x\mapsto (x-a)^2$ and $x\mapsto (x+a)^2$. Thus, if $x=0$ doesn't work, try another $x$ :) Sep 17, 2015 at 11:08
• Yep it says so now, anyway ;) Sep 17, 2015 at 11:09

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

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

so, the $$a=0$$

• You are only allowed to divide by $(x+a)^2$ if it is not $0$. Sep 17, 2015 at 8:08
• Also the value inside the square can be $-1$. Sep 17, 2015 at 8:22
• @mathreadler yes you are right ,this case is happened when $x=0$ Sep 17, 2015 at 8:32

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.

• This is true for any field of characteristic $2$. Sep 17, 2015 at 6:53

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