### September 9, 2003

I found a very cool math problem in my math book today. It looks scary, but it's not. I use the quadratic formula in there, so check my Graphs & Functions page for a refresher, if you need it.

$$(x^2 + 3x)(\sqrt{x + 1} - 4) = 0$$

Find values of $$x$$.

First of all, let's figure out what $$x$$ can't be. We know that it's illegal to take the square root of a negative number, so $$x > -1$$. As you can see, there are two factors in this equation, and they equal $$0$$, which means that one of them (or both) have to be zero. So we check them both:

$$x^2 + 3x$$

$$x_1 = 0$$ $$x_2 = -3$$

$$-3$$ is a false answer, since $$x$$ has to be greater than $$-1$$. We check the second factor:

$$\sqrt{x + 1} - 4 = 0$$ $$\sqrt{x + 1} = 4$$ $$x + 1 = 16$$ $$x = 15$$

Now we just have to insert the two values $$0$$ and $$15$$ into the original equation, and see which of them yields $$0$$ as an answer. You can do that for yourself. :-)