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Math .. Rationalize the Denominator











































Rationalize a denominator containing three terms.

1/1+rad3-rad5

You can apply the same reasoning to rationalize a denominator containing three terms as you would use to rationalize a denominator containing only two terms.

The key is to group the terms.

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Jun 20, 2011
Rationalize a 3 term Denominator
by: Staff


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Part II



Repeat the process - Multiply by a fraction containing the new conjugate: (2*3^1/2 + 1) in BOTH the numerator and denominator.


(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)]


As before, multiply both numerators and multiply both denominators, just as you would when multiplying any two fractions:


= [(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 - 1²)

[(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




Check this answer against the original expression with a calculator:

Final answer:

(7 + 3 * 3^1/2 + 5^1/2 + 2 * 15^1/2) / 11 = 2.0162

Original expression:

1/(1+3^1/2-5^1/2) = 2.0162




Thanks for writing.

Staff
www.solving-math-problems.com



Jun 20, 2011
Rationalize a 3 term Denominator
by: Staff


Part I


The question:

1/1+rad3-rad5


The answer:


Rationalizing the Denominator is the standard way of simplifying fractions containing radicals in the denominator. Rationalizing the denominator means to “rewrite the fraction so there are no radicals in the denominator”.


Your problem as it now stands:


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


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


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

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


Multiply the original fraction by a fraction containing the conjugate: [(1 + 3^1/2) + 5^1/2] in BOTH the numerator and denominator.

The fraction [(1 + 3^1/2) + 5^1/2] / [(1 + 3^1/2) + 5^1/2] is equal to 1, so the original fraction is merely being multiplied by 1. As you can see by the following illustration, its value has not been changed.


= [original fraction]

= [original fraction] * [(1 + 3^1/2) + 5^1/2] / [(1 + 3^1/2) + 5^1/2]

= [original fraction] * 1

= [original fraction]

Therefore,


= [original fraction] * [(1 + 3^1/2) + 5^1/2] / [(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]}

Multiply both numerators and multiply both denominators, just as you would when multiplying any two fractions:

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 * [(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)

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