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Rationalize a 3 term Denominator

by Asia
(Las Vegas)

Rationalize a 3 term Denominator

Rationalize a 3 term Denominator











































Rationalize a Denominator containing 3 terms

The difference of squares formula states that:

(a + b)(a − b) = a^2 − b^2

You can apply the same reasoning to rationalize a denominator which contains three terms by grouping the terms.

Rationalize the denominator of the following expression.

1/(1+3^1/2-5^1/2)

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Feb 15, 2011
Rationalize a 3 term Denominator
by: Staff

The question:

by Asia
(Las Vegas)

1/(1+3^1/2-5^1/2)



The answer:

Your problem has three terms in the denominator: a + b + c

However, imagine for a moment how you would rationalize a denominator with only two terms: a + b.

As you know, if the denominator contains only two terms, you could rationalize the denominator by multiplying the denominator by its conjugate: a - b.

The difference of squares formula states that:

(a + b)(a − b) = a^2 − b^2


You can apply the same reasoning to rationalize a denominator which contains three terms. GROUP THE TERMS as follows:

a + b + c = (a + b) + c

The difference of squares formula states that:

[(a + b) + c] * [(a + b) - c] = (a + b) ^2 − c^2


Original Problem:

1/(1+3^1/2-5^1/2)

Group the denominator so that the difference of squares formula can be applied:

1/[(1 + 3^1/2) - 5^1/2]


Multiply by the conjugate: [(1 + 3^1/2) + 5^1/2]

1/[(1 + 3^1/2) - 5^1/2]

{1/[(1 + 3^1/2) - 5^1/2]} * {[(1 + 3^1/2) + 5^1/2] / [(1 + 3^1/2) + 5^1/2]}

(1 + 3^1/2 + 5^1/2) / [(1 + 3^1/2)^2 + (5^1/2)^2]

(1 + 3^1/2 + 5^1/2) / (1 + 2*3^1/2 +3 ? 5)

(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)


Repeat the process - Multiply by the new conjugate: (2*3^1/2 + 1)

(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)

[(1 + 3^1/2 + 5^1/2) / (2*3^1/2 ? 1)] * [(2*3^1/2 + 1)/ (2*3^1/2 + 1)]

[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / [(2*3^1/2 ? 1) * (2*3^1/2 + 1)]

[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / (2*3^1/2)^2 ? 1^2)

[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / (12 - 1)

[(1 + 3^1/2 + 5^1/2) * (2*3^1/2 + 1)] / 11

(2 * 3^1/2 + 6 + 2 * 15^1/2 + 1 + 3^1/2 + 5^1/2) / 11

(7 + 3 * 3^1/2 + 5^1/2 + 2 * 15^1/2) / 11


The final answer is: (7 + 3 * 3^1/2 + 5^1/2 + 2 * 15^1/2) / 11



Thanks for writing.


Staff
www.solving-math-problems.com



Sep 06, 2016
rationalixe the denominator NEW
by: Anonymous

i ask rationalize the denominaor with three terms and diffrents coeffients

Jun 03, 2017
Plz tell how to solve the given ques NEW
by: Anonymous

√3-√4 upon √3+√4-√7

Oct 06, 2021
Math Format
by: Anonymous

I'm not gonna lie, but the format the answer was written in was a royal pain in the a** to read. I couldn't even copy and paste it to WolframAlpha or some other math site to make it legible because the format was either incompatible or the order of operations was incorrect. This is very annoying as this is the only material I could find online regarding rationalizing any denominator with more than two terms.

Nov 03, 2021
How to Rationalize (√3-√4)/(√3+√4-√7)
by: Unbounded Intellect

Hi, I’m the same guy who posted above about this defunct site’s abysmal formatting. I know it’s been a long time since the last comment asked this question, but I’ll answer it anyway in case there is anyone else who happens to stumble upon this obscure page and still has trouble understanding this. Also so I can answer using slightly better formatting and actually using the symbols √ and ² instead of ^½ and ^2 so this isn’t an eyesore to read.

The fully rationalized and simplified answer is (3√7−2√21−√3)/12. In decimal, this is approximately equal to −0.2466623...

First thing to do is simplify √4 to 2. So the question becomes: (√3−2)/(√3+2−√7)

Next, you can add parenthesis and group the first two terms of the denominator (√3+2) to make solving easier. (√3−2)/[(√3+2)−√7)]

Then you multiply both the numerator and the denominator by the conjugate of the denominator as usual. The conjugate of [(√3+2)−√7)] would be [(√3+2)+√7)].

So you have (√3−2)/[(√3+2)−√7)]*[(√3+2)+√7)]/[(√3+2)+√7)] →
(√3−2)*[(√3+2)+√7)]/[(√3+2)−√7)]*[(√3+2)+√7)]

Let’s multiply out the numerator first. (√3−2)*[(√3+2)+√7)] is in the form: (a−b)[(a+b)+c]
That distributes to a²−b²+ac−bc. Which is the difference of squares for 'a' and 'b' plus 'c' distributed to the first binomial.

So after substituting that would make it (√3)²−2²+√3√7−2√7. Which simplifies to
3−4+√21−2√7 → √21−2√7−1.

For the denominator, [(√3+2)−√7)]*[(√3+2)+√7)] is in the form: [(a+b)−c][(a+b)+c]. Another way to write this is (a+b)²−c². This distributes to (a²+2ab+b²)−c². This is the difference of squares of (a+b) and 'c' if you treat (a+b) as a single variable. This is the benefit of grouping like terms.

After substitution: [(√3)²+2(√3)(2)+2²]−(√7)² simplifies to (3+4√3+4)−7 → 4√3.

So the expression after rationalizing the denominator once is (√21−2√7−1)/(4√3). But we’re not done yet. There is still the radical √3 remaining in the denominator. So we have to rationalize it a second time, albeit this time is much easier.

We multiply the numerator and denominator by √3/√3 so you have √3*(√21−2√7−1)/√3*(4√3).

Let’s get the numerator over with. √3√21−2√3√7−√3 → √63−2√21−√3. The radical √63 can be further simplified using prime factorization. Prime factorization, in case you aren’t familiar with it is when you decompose a number into its prime factors by starting with dividing by the lowest prime number 2, and dividing by higher primes if the number is not perfectly divisible by the lower ones.

Since 63 is an odd number and therefore not divisible by 2, the next highest prime number that it is divisible by is 3. 63/3 is 21, which in turn is divisible by 3 to get 7. So the prime factorization of 63 is 7, 3, and 3. Notice how there are two 3's. Another way to write the factors is 7 and 3² which equals 9. 9 can be square rooted back to 3. So √63=√(9*7)=3√7.

So the most simplified numerator is 3√7−2√21−√3.

Now let’s work on the denominator, which is very easy to rationalize. √3*(4√3) → 4√3√3 → 4*3=12. That was fast.

So the final answer is (3√7−2√21−√3)/12. Hope this helped anyone who happened to read this.

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