This little puzzle popped up the other day and I thought it would make a good, simple example of how to deal with square roots.

First, a little terminology: The word for *root* in latin is *radix* from which we get words like *radius *and *radical*. Mathematicians refer to the square root symbol √ as the *radical* symbol, and the stuff we’re taking the square root of they’ll refer to as what’s *under the radical*.

We start with this.

We can get rid of square root symbols by just squaring what’s under them, which will leave just the terms *under the radical* and nothing else. Since this is an equation, we’ll need to do this to both sides:

When we multiply this out, it looks like a real mess:

We can combine the two x’s to make this

But, it isn’t all that much simpler. Let’s try another approach. Having two different radicals on the same side of the equation is what makes this messy, so, let’s move one of them to the other side of the equals sign. We can move the term by subtracting it from both sides.

Now let’s square both sides of the equation like we did before and see what we get:

I know, it doesn’t look all that much simpler, but, notice how there is a solitary x term on both sides? We can get rid of that by subtracting x from both sides:

leaving us with

It’s looking easier already! Let’s subtract 15 from both sides, and then add to both sides. We get

Now we have simple terms for both left and right hand sides of the equation. Divide both sides by 30 to get

Square both sides (yes, *again) *to get rid of the radical, and we end up with

Let’s see if this is right. Plug 49 in for x in the original equation:

We can add the 49 and 15 under the first *radical* to get

The square root of 64 is 8 (8 x 8 = 64) and we already have square root of 49 being 7, so

So, the answer checks out!

Totally *RADICAL*, eh?