# Verifying solutions to equation

There is an example of an equation in a text book I've looked at which puzzles me slightly:

We have:

$\sqrt{x+2} = x$

$\Downarrow$

$(\sqrt{x+2})^2 = x^2$

$x+2=x^2$

$x^2-x-2=0$

$(x-2)(x+1)=0$

$x=-1$ or $x=2$

We then have to verify that the solutions are correct. For $x=2$, we get:

Left side = $\sqrt{x+2} = \sqrt{2+2} = 2$

Right side = $x=2$.

We see that the solution is valid.

For $x=-1$, we get:

Left side = $\sqrt{x+2} = \sqrt{-1+2} = 1$

Right side = $x = -1$

The left side is not equal to the right side, so the solution is not valid.

My question is this: In the latter verification we have, for the left side, $\sqrt{x+2} = \sqrt{-1+2} = \sqrt{1}$. But why is it only that the positive square root of $1$ is used in the verification? After all, $\sqrt{1} = \pm 1$, and if we include the negative solution, then the two sides are equal.

If anyone can explain to me why we do not include the negative solution here, then I would greatly appreciate it!

• It's a worldwide agreement amont mathematicians: to avoid misunderstandings and confusion, we always choose the positive root of a non-negative real number nless otherwise specified. With complex numbers things become messier, but again: it is a matter of choosing something.. – DonAntonio Sep 12 '16 at 19:06
• @Roby5 I think that's precisely the question of the OP: why? – DonAntonio Sep 12 '16 at 19:06

## 2 Answers

By definition $\sqrt{a}$ is the positive number x such that $x^2=a$. So, if $\sqrt{x+2}=x$, $x$ must be $\ge 0$.

In other words: a correct solution of the equation $\sqrt{x+2}=x$ require the solution of the system: $$\begin{cases} x+2=x^2\\ x\ge 0 \end{cases}$$

• Thanks a lot! This explains it very well :). – Kristian Sep 12 '16 at 19:26

After all, $\sqrt{1} = \pm 1,\dots$

This is not correct. These two are correct:

$$\sqrt1 = 1$$ $$-\sqrt1 = -1$$

When we say things like $\sqrt{x}$, where $x$ is positive, we just mean the positive square root. It's a convention. (Yes, we can take the square root of zero, but $\sqrt0 = 0$ is neither positive nor negative, which is why I'm restricting $x > 0$ for the sake of this discussion.)

On a related note, it is true that both $\pm1$ satisfy $x^2 = 1$, which is why when we solve $x^2 = 1$ we have $x = \pm 1$. But if we want all $x$ such that $x = \sqrt1$, then only $x=1$ works. (And likewise if we want all $x$ such that $x = -\sqrt1$, then only $x=-1$ works.)

• Well, one could also always choose $\;\sqrt1=-1\;,\;\;\sqrt4=-2\;$ , etc. and agree on that. True, that'd cause another kinds of problems in other parts of the now agreed on mathematics, but there's nothing wrong with taking the negative root per se. The restricition $\;x\ge0\;$ is necessary in the real numbers, not merely to make things simpler. – DonAntonio Sep 12 '16 at 19:08
• @DonAntonio, yes, but the OP is clearly at a precalculus level and I don't think such concepts are helpful for questions asked at that level. – tilper Sep 12 '16 at 19:11
• Thanks a lot for your input! This is much clearer to me now :). – Kristian Sep 12 '16 at 19:26