# How is the plus-minus sign used for solving exponential and radical equations?

I noticed that for an equation such as $(4x+1)^2 = 289$, the plus-minus sign $\pm$ is used right after eliminating the squared expression $(4x+1)^2$ via square rooting.

Yet, for an equation such as $\sqrt {4x-1}$ $=$ $\sqrt{x+2} - 3$, the plus-minus sign $\pm$ is not used right after eliminating the radical expression $\sqrt{4x-1}$ via squaring.

So, why is the $\pm$ sign introduced right after taking the square root of a square, but not right after taking the square of of a square root?

• Because one is square-rooting and one is squaring. A number can have two different square roots, but can only have one square. – peterwhy Dec 23 '14 at 2:20

This is because the function that turns $x$ into $x^2$ is not a one-to-one function, which simply means that different numbers sometimes have the same square. For example, $7^2 = (-7)^2 = 49$.

As a result, if you know $x^2=y^2$, you can only conclude that either $x=y$ or that $x=-y$. You do not know which it is, because there are two different numbers that have the same square. We abbreviate this situation by writing $x=\pm y$.

But if $x=y$, then $x$ and $y$ are equal, so they have the same square, and you can conclude $x^2=y^2$; no $\pm$ is necessary because there is only one possibility.

Similar situations apply for all sorts of functions that are not one-to-one:

• Consider the function $f(x) = 0$ if $x$ is even, but $f(x) = 1$ if $x$ is odd. Then if you know $f(x) = f(y)$ you can only conclude that $x$ and $y$ have the same parity: either both are even or both are odd. We abbreviate this situation in a special way, and write "$x\equiv y \pmod 2$". But if you know $x=y$ you can still conclude $f(x) = f(y)$: $x$ and $y$ are the same, so their $f$'s are the same also.
• If you know $e^x = e^y$ and $x$ and $y$ are real numbers, you can conclude $x=y$, because the function $e^x$ is one-to-one on the real numbers. It is not one-to-one on the complex numbers, so if $x$ and $y$ might be complex numbers, you cannot conclude $x=y$; you can only conclude $x = y+2ki\pi$ for some real integer $k$. But if you know $x=y$, you can still conclude $e^x = e^y$.
• If you know $x\cdot 0 = y\cdot 0$ you can conclude nothing at all about $x$ and $y$. (We don't have a special notation for this because it would be useless.) But if you know $x=y$ you can certainly conclude $x\cdot 0 = y\cdot 0$!

This is too long to put as a comment so I'm leaving it down here instead. @MJD posted a good answer above mine already.

It should be mentioned as well that the two functions $f(x) = x^{\frac{1}{2}}$ and $g(x) = \sqrt{x}$ are very much related. To the casual observer, they appear to be the same thing, however they are infact slightly different.

$f(4) = 4^\frac{1}{2} = \pm 2$

$g(4) = \sqrt{4} = 2$

The reason is that with a fractional exponent (or non-integer exponent in general), it is a multivalued function, a map that is not well-defined and has multiple potential answers.

In the case of using a $\sqrt{~~}$ or similar, we are interested only in the principle square root, i.e., the positive real solution to $x^\frac{1}{2}$.

When converting an equation such as $x^2 = y^2$, although we often say "squareroot each side", what we are infact doing is raising each side to the power of $\frac{1}{2}$, hence the plus-minus.