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?
Well, your first step right out of the gate is problematic (as is your first equation, but we'll get to that in a moment).
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?)