Math: simplifying radicals, Part II

Tutoring math, you need to explain this every semester.  The math tutor draws the distinction between handling the numbers and the variables.

Looking back to my earlier post on radicals, you’ll get the basic idea. Just for a quick review, suppose you have √50. You need to break it down to √25√2. Then, take the square root of 25 to get 5. Your simplified answer is 5√2.

Note that you need to break the number into a perfect square times whatever else. If a perfect square doesn’t factor into the number, you can’t simplify it. For example, √35 does not simplify. You can write it as √7√5, but neither 7 nor 5 is rootable (into a whole number). Therefore, we leave it √35.

What about √x10? Well, the answer is x5. The reason is that (x5)(x5)=x10. When you multiply the two terms, you add their exponents. Therefore, √x14=x7.

Now, let’s look at √x15. How do we simplify it? We break it down as √x14√x. Now we realize that √x14=x7. Our simplified answer is x7√x.

Next time, we’ll tackle a situation with both numbers and variables.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Leave a Reply