# Dealing with negatives in radical

I have this question in my homework and I'm unable to find what $x$ is. I keep coming up with a negative number but I don't think it can be a negative number because you can't have $\sqrt{-x}$. Here's the question:

Solve the following equation. State any restrictions on x.

$\sqrt{3x+15} = -6$

Here was my attempt at it:

$$\sqrt{3x+15} = -6^2$$

$$(3x)^2+15^2 = 36$$ $$9x^2 + 225 = 36$$ $$9x^2 + 225 - 225 = 36 - 225$$ $$9x^2 = -189$$ $$\frac{9x^2}{9} = \frac{-189}{9}$$ $$x^2 = -21$$ $$x = \sqrt{-21}$$

Now if I go backwards and try to replace $x$ with $\sqrt{-21}$ this is what happens:

$$\sqrt{3(\sqrt{-21}) + 15} = NaN$$

I understand it's because of $\sqrt{-21}$ and you can't find the squareroot of a negative number, but that means there's another solution to this. Any ideas?

• When you square $\sqrt{x+3}$ you should get $x+3$. And the whole question is rather strange, since usually $\sqrt{b}$ is defined to be the non-negative number whose square is $b$. – André Nicolas Jan 14 '14 at 0:52

If $x$ is a number such that $$\sqrt{3x+15}=-6,$$ then we have $$3x+15=(-6)^2.$$ This is because for any real number $b,$ $\sqrt{b}$ is necessarily a number whose square is $b.$ At that point, we have $$3x+15=36\\3x=21\\x=7.$$ However, when we resubstitute, we find that $3x+15=36$ (as we might expect), but $\sqrt{36}=6\ne-6.$ This is because $\sqrt{b}$ is defined to be the nonnegative number (if any) whose square is $b.$ If we keep that in mind, we should see that something is wrong at the very start, and that the equation has no real solution.
As a side note, $\sqrt{-x}$ does make sense if we happen to know that $x$ is a nonpositive real number. Also, $\sqrt{3x+15}$ makes sense (in the context of real numbers) exactly when $x\ge-5.$ (Why?)
$$(\sqrt{3x+15})^2 = -6^2$$
$$3x + 15 = 36~~~~\text{ Square from both sides}$$ $$3x = 21~~~~~~ \text{ Subtract 15 from both sides}$$ $$x = 7~~~ \text{ Divide both sides by 3}$$
One key thing to note is that when you square a square root, you do not distribute the square inside. $(\sqrt{a+b})^2= a+b$. This is not (usually) the same as $a^2 + b^2$.