﻿ Archive — hermiene.net

# Archive — hermiene.net

"You have enemies? Good. That means you've stood up for something, sometime in your life."

### 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. :-)