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. :-)