Radical Algebra Variables in Laguna Niguel
On a recent Algebra tutor blog post, I wrote about “radical” Algebra problems with square roots. As an experienced Algebra tutor, I always say the most important thing for radical Algebra problems is that you break down the radicand (that number inside the radical) into at least one perfect square.
Radical Algebra Integer Review
So, if you have the problem √40, you can break this radical expression down into √(4×10) or √4√10. Since √4 is a perfect square, its perfect square 2 “breaks out” of “radical prison,” leaving the imperfect square 10 forever stuck in radical prison. Thus, √40 simplifies to 2√10.
What do you do, though, when the radicand includes variables? Many Algebra students in Laguna Niguel have an easy enough time simplifying radicands with integers, but they struggle with radicands that include variables. The process of breaking variables out of radical prison, however, works just the same as breaking out integers.
You just need to know your Algebra variable exponent rules. Here’s a quick review.
Radical Algebra Exponent Review
When taking the square root of a variable with an exponent, it’s important to remember Algebra exponent rules. Take the following variable expression: x6
This simple expression could be broken down several ways. x5x1, x4x2, and x3x3 are all valid rewrites of the expression x6. The key here is that anytime multiple variables of the same base (in this case, x) are multiplied together, you add the exponents. In this problem, 5+1 and 4+2 and 3+3 all equal 6, so any combination of those added exponents with the same multiplied base x will produce the variable expression x6.
Thus, you could also rewrite x8 as x6x2, x5x3, x4x4, and so on. Once you know the exponent rules of Algebra, figuring out how to simplify a radical expression is a piece of cake!
Radical Algebra Simplification Review
Let’s return to our original variable expression example, x6. Only now, let’s throw it in radical prison: √x6
The quick and easy way to simplify this expression is to break it down into its perfect squares. In this case, we already know x6 simplifies to x3x3. It’d be like the number 4 breaking down into 2(2). Since the perfect square of x6 is x3, x3 is what breaks out of radical prison, leaving nothing left inside the radicand.
Obviously, there are more difficult examples, but always start with breaking down your variables into a perfect square! If you had √x9, you could break the radicand into the following: √x4x4x1. Since we have two x4, we can break out one x4 from radical prison, leaving a lonely little x still trapped inside the radical: x4√x
If you can find the perfect squares for your radicand, you can do the rest. Have a blast breaking integers and variables out of radical prison!