# Radical Radicals

Today’s math student does not like radicals any more than yesterday’s student does. Although he or she is armed with a calculator, the guessing processes of how to break down radicals or when to rationalize the denominator still dominate the thinking and strategy processes. As well, students are not always sure that their final answer is in lowest terms, especially when dividing. There has to be a set approach to all operations involving radicals that the student can use to resolve some of these issues.

Consider an example like . We do suggest to our students that they reduce the radicands before combining terms. The logical question, of course, is whether these terms can be combined to begin with, and if so, how to proceed with the problem. In this case, how is the innumerate student supposed to know that 3 is the number to start with, especially if we have taught our students to extract perfect squares from each of the given radicals? How is he or she to proceed with confidence and certainty to begin with?

We then state that we cannot add or subtract the terms if the g.c.f. of the radicands is 1. We also suggest not to multiply or divide until we have determined the g.c.f. of the given radicands. As well, we do not rationalize the denominator unless the g.c.f. of both numerator and denominator is 1.

The trained eye observes that g.c.g. (27,75) is not 1 but 3: thus, the question is now worth doing. As 3 is a common factor of both 75 and 27, we now express and in terms of . Instead of having to guess how to break down both 75 and 27, we already have one of the key factors. The student then has the simple task of dividing 3 into 27 and 75 before simplifying the terms.

Consider the conventional way of multiplying two radicals: e.g., . Many calculator-oriented students would multiply 28×65=1764 and then try to break down only to find 42. If we now tell the student to break them down first before multiplying, a guessing or guess/estimating process begins as to what numbers go into 28 and then into 63. It is discovered that and that , but only after calculating the numbers one at a time.

The process of Radical Radicals involves considering both radicands at the same time by first finding their g.c.f. To do this, we use Euclid’s Algorithm (without calculator).

## (2)

e.g. Find g.c.f (28,63)

Steps

- Divide the larger number by the smaller number, keeping track of the remainder.
- Now divide the remainder into the previous divisor.
- Continue step 2 until the remainder is exactly zero.
- The divisor that yields a remainder of zero is our g.c.f.

____2___ ____4___

28) 63 7) 28

__56____28__

7.0

We know 28 and 63 have 7 as a greatest common factor and so and have as a common factor. This brings about a different way of doing radicals because we work with two radicands at a time, not one. As well, when it does come time to break down the radicals, half the work is already done, as we already know on of the factors. Thus, guessing of guess/estimating is reduced dramatically.

Now consider . Instead of looking at , we determine that g.c.f. (26,65) = 13 and express each factor in terms of .

This gives

=

=

The numbers never become large, so simplifying is easier.

The disadvantage of this method is that students must find the g.c.f. of two numbers with little help from a calculator. The advantages are that once the g.c.f. has been found, half the work is already done because guessing is virtually eliminated and the numbers never increase in size. This reduces frustration and error made with large numbers.

In summary, the steps to this new approach are as follows:

- Find the g.c.f. of the radicands
- Express each radicand in terms of the square root of this radicand.
- “Pair-off” like radicals.
- Simplify remaining terms.

## (3)

If you think multiplication done in this fashion is efficient, let us now look at division. This is where this method really shines.

Consider .

Conventionally,if the student does not recognize that 7 is a common factor, he/she might multiply top and bottom by , giving horrendous numbers to work with.

Instead, use Euclid’s Algorithm to find that g.c.f. (112,175)=7. Thus, we have

This leaves the answer , with no other calculations to do.

What happens if the g.c.f. of the numerator and denominator is 1? We simply multiply top and bottom by the denominator knowing that we will not have to reduce the fraction after the multiplication of terms.

Now, what happens if the numerator is not a radical? Consider .

Here we do not multiply top and bottom by because g.c.f. (2,10)=2 (not 1).

Instead we recall that and in this case . Since g.c.f. (1,100)=2, we express both numerator and denominator in terms of . This gives

The “cancels,” and we are now left with . Since g.c.f. (2,5) = 1, we can now multiply top and bottom by and not have to worry about reducing the new function.

Finally, we have

## (4)

We do not have to go backwards or look over our shoulder to see if the fraction can be reduced as the g.c.f. is 1. Our rule then is to only multiply top and bottom by the denominator factor when g.c.f. is l.

In general, the steps for division are as follows.

Step 1. Find the g.c.f. of both numerator and denominator.

Step 2. Express each radicand in terms of the square root of this g.c.f.

Step 3. “Pair-off” like radicals and reduce.

Step 4. Simplify remaining terms.

Step 5. Multiply top and bottom by the denominator term only when the g.c.f. of numerator and denominator is 1.

For division then, we merely add step 5 to the method for multiplication.

This new process now simplifies addition and subtraction considerably because we can operate only if the g.c.f. is not 1. Hence, for , we suggest that this cannot be simplified any further because g.c.f. (3,7)=1. By the same token, for , we find that g.c.f. (24,54)=6. We then have

=

=

=.

Again, we look at two at a time, not one at a time. We then ascertain that terms can indeed be combined because their g.c.f. is not 1. If indeed we have more than two terms, we can then look for two that have the same g.c.f.

For combined operations, we look at 2 g.c.f.’s as opposed to 1. Consider an example like . Since g.c.f. (3,15)=3 and g.c.f. (3,30)=3, we have and . Thus, we can write

=

=.

## (5)

For division, take an example like . Now g.c.f. (5,7,2)=1, so we do multiply top and bottom by the conjugate .

This gives

=

=

=.

In summary, by finding the g.c.f. of the radicands we introduce the idea of working with two radicands at a time, not one at a time. And, once we do have the g.c.f., we cut the work by a least half because we have one of the factors of the radicand. We eliminate large numbers, errors caused by large numbers, and the frustration from guessing. In a proactive way, we streamline the process by keeping numbers simple and neat. In my experience, student feedback is always the same: “This is easy compared to what I used to do.”