Tutoring high school math, you realize that radicals are a problem for many. Today the math tutor gives brief coverage of multiplying radicals.

Consider the following example:

Simplify 3√(6)*4√(7)

Multiplying with radicals goes “number times number, radical times radical.” Like so:

Step 1: Number times number: In our case, 3 times 4, which gives 12.

Step 2: Radical times radical. In our case, √(6)√(7) gives √(42).

Putting it all together, we get

3√(6)*4√(7)=12√(42)

So far, so good. Beware, though: a simplification can creep in when the radicals are multiplied together:

Example 2: Simplify 3√(10)*7√(6)

Solution: Using number times number, radical times radical, we get

21√(60)

From my earlier post on simplifying radicals, we know we can’t just leave the answer 21√(60). After all, 60 contains the perfect square 4; we must simplify it as follows:

21√(60)=21√(4)√(15)=21*2√(15)=42√(15)

Note that if one of the factors is missing the number part, you just imagine it’s a 1 and proceed:

4√(7)*√(11)=4√(7)*1√(11)=4√(77)

On the other hand, one of the factors might be missing the radical part. Then, you just multiply the numbers together and tack the radical at the end:

6√(5)*(8)=48√(5)

Visit again for more tips on operations with radicals:)

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