Rationalize Denominator With Square Root

I of the typical problems encountered in elementary algebra is rationalizing the denominator. Remember that rationalizing the denominator means to catechumen the original fraction that contains at least i square root term in the denominator into a new fraction, of equal value, that does not have a foursquare root term in the denominator. This article will take the reader through several simple steps that bear witness how to rationalize the denominator for virtually common fraction types. There is, also, a formula department below each example that provides a detailed algebraic method used in deriving a formula wherein one tin can simply plug in the values of the original fraction into a new fraction with a rational denominator.

Equally the reader progresses through the vii major methods and their resulting final formulas, 1 will chop-chop discover that each subsequent fraction type volition become more complicated to rationalize and the resulting fraction with a rational denominator more complicated in turn.

The mathematical level involved in this article is simple elementary algebra; however, as the fractions become more difficult, the steps will take longer to solve and result in tedious computations. With that being said, let us start with a simple fraction and progress forward:

#one) Rationalizing the denominator with a single square root in the denominator:

Given i/√2 how does one solve it? The answer to solving every trouble that contains a denominator that needs to be rationalized is to multiply the unabridged fraction by the equivalent value of 1. Remember when a fraction is multiplied past one, the value of the fraction does non change. What changes is how the fraction "looks." For example solving the above fraction we take:

Every bit can be seen past the above example, the fraction was multiplied by the value of the existing denominator in the original fraction. In order not to modify the value of the fraction, the fraction needed to be multiplied by the square root of 2 in both the denominator and numerator. This second fraction, called a modifying fraction, in one case multiplied to the original fraction converts the denominator of the original fraction into a term that does not contain a radical term (i.e. a square root term). The end outcome is a new fraction with the denominator rationalized while nonetheless maintaining the original value of the unmodified fraction.

This is the basic technique that volition be applied over and once more to each type of fraction presented. The mathematical variations used to solve each subsequent trouble will exist slight.

The formula section volition particular the algebraic method used to derive a simple plug-in formula used to rationalize these types of fractions. These formulas are useful to have handy especially if yous will be encountering many such bug and do not have the time to rationalize each and every fraction encountered.

Formula for equation #1:

#2) Rationalizing the denominator of a cubic, or higher, root.

The fox hither is to realize that one must multiply the initial fraction in such a manner that the denominator has been completely rationalized. For instance: If the denominator is a cubic root, root three, the fraction needs to be multiplied by itself twice. If the denominator is a 10th root, root x, then it would need to be multiplied by itself nine times. In brusque, whatever value the root is, root "c," the fraction needs to be multiplied (top and lesser) by itself one less times the value of root of the original fraction. That is it must be multiplied by itself root "c-1" times.

In this case the root has a value of ix so it needs to be multiplied by itself 8 times since (9 – 1 = 8).

Formula for equation #2:

#3) Rationalizing the denominator of a cubic, or college, root that contains a squared, or higher, term inside the root:

This is a slight variation of the in a higher place instance. Since the original fraction has a term that raised to a higher power than ane, all one needs to exercise is find the departure betwixt the power inside the radical and the root of the radical and use that departure constitute to multiply the modifying fraction by that deviation.

For case: If the fraction has is fifth root and the term inside is a cubic, raised the third power, and then the difference between five and 3 is two. That is: 5 - 3 = 2. See example beneath:

Formula for equation #3 for c > d:

#4) Solving fractions that incorporate two terms in the denominator: with 1 term being a foursquare root.

To solve these types of fractions 1 needs to innovate the concept of the conjugate. The cohabit is a elementary mathematical construct. A conjugate is simply the opposite of the improver or subtraction operation used when combining two terms. For case if you accept 5 + 3 as the original set of terms then the conjugate would be written as 5 – 3. It is that simple! The advantage of multiplying a set up of terms by its cohabit is that the resulting set of terms will exist the divergence of 2 squares. For example: (five+iii) x (5-three)=(5x5 + (5x3 -5x3 this cancels)- 3x3).

Why would this be advantageous one might enquire? The reason is simple: One wants to find a method to multiply the square root by itself thus eliminating the square root term!

Therefore the purpose of multiplying by the conjugate is to eliminate the foursquare root. In all problems you will desire the square root to be removed from the denominator of the fraction. If y'all start with a "+" between the two numbers you multiply the same two number with "-" between them and vice versa.

The example below shows how to solve a fraction using a conjugate when you have a two terms one of which is foursquare root in the denominator:

Formula for equation #4:

#five) Using a conjugate when yous have two square roots in the denominator:

Formula for equation #4:

#6) Using a conjugate when you have three terms in the denominator which two of them are two separate foursquare roots:

In solving this trouble, group any two terms as if they were a single term. Then multiply the conjugate betwixt the two terms acting as ane and the 3rd term in the original fraction. See case beneath:

Step 1: Multiply by the conjugate (note the first two terms were grouped together):

Pace 2: Expand the term that is squared and consolidate the resulting results:

Step 3: Multiply by the cohabit again:

Formula for equation #five:

Expanding the denominator:

Using the conjugate again we get formula for equation #half dozen:

#7) Rationalizing the denominator when it has 3 square root terms:

Pace i: Multiply by the cohabit (once again selection two terms to act as ane):

Step two: Aggrandize the term that is squared and consolidate the results:

Step iii: Multiply by the conjugate once more:

Deriving the formula:

Expanding the denominator:

Multiply by the cohabit once again we get equation #7:

As can be seen in the above vii examples, each resulting fraction becomes more and more than complicated. The question that students ask is why get through all that trouble when the initial fraction appears to be simpler to use versus the resulting fraction that has been rationalized. A good questions indeed. The answer lies with precision. Before the advent of computers, all computations were washed manually. The rule of thumb was to practise equally many "precise" computations offset then truncate the result. In nearly cases, performing the radical functioning, results in a number that is infinitely long. If you split up a number that is infinitely long, one needs to truncate information technology first so perform the division. If that aforementioned number were in the numerator then y'all would carve up past a fixed length number into that radical term, perform the root, and then truncate the result.

That is why in that location was a demand to rationalize the denominator. Today it has become function of mathematical convention and the irritation of millions of algebra students.

Summery equations and their multiplying identities are listed below: