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rationalize denominator

by ann
(los angeles)











































simplify the following expression by rationalizing the denominator

square root of two divided by 3-square root of 6

the square root of 6 in the denominator is an irrational number.

rationalizing the denominator will ensure that there will not be an irrational number in the denominator.

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Jan 30, 2011
Rationalize Denominator
by: Staff


The question:

by Ann
(Los Angeles, CA, USA)

square root of two divided by 3-square root of 6




The answer:

square root of two divided by 3-square root of 6

sqrt(2)/[3*sqrt(6)]

notice that the square root of 6 can also be written as follows:

sqrt(6) = sqrt(2*3)

sqrt(2*3) = sqrt(2) * sqrt(3)

sqrt(6) = sqrt(2) * sqrt(3)


because the sqrt(6) can be written as sqrt(2) * sqrt(3), your expression can be rewritten as:

sqrt(2)/[3 * sqrt(2) * sqrt(3)]

this expression can be simplified further by rewriting it as two fractions:

sqrt(2)/[3 * sqrt(2) * sqrt(3)]

[sqrt(2)/ sqrt(2)] * {1/[3*sqrt(3)]}

Notice that the numerator and denominator of the 1st fraction [sqrt(2)/ sqrt(2)] are the same. This means they will cancel one another. The 1st fraction = 1:

[sqrt(2)/ sqrt(2)] * {1/[3*sqrt(3)]}

[1] * {1/[3*sqrt(3)]}

1/[3*sqrt(3)]


The last step is to remove the sqrt(3) from the denominator of the simplified expression

We can do this by multiplying the expression 1/[3*sqrt(3)] by 1. We will not use an ordinary 1. Instead, we will use the fraction sqrt(3)/ sqrt(3). This fraction does = 1.

1 = sqrt(3)/ sqrt(3)



{1/[3*sqrt(3)]} * 1

{1/[3*sqrt(3)]} * [sqrt(3)/ sqrt(3)]

This expression can be rewritten as the following two fractions.

{1/[sqrt(3)*sqrt(3)]} * [sqrt(3)/ 3]

The product of sqrt(3)*sqrt(3) is simply 3.

Therefore:

{1/[sqrt(3)*sqrt(3)]} * [sqrt(3)/ 3]

{1/3} * [sqrt(3)/ 3]

Multiplying the two fractions:

{1/3} * [sqrt(3)/ 3]

[1 * sqrt(3)]/(3 * 3)

[sqrt(3)]/(3 * 3)

sqrt(3)/(3 * 3)

sqrt(3)/(9)

sqrt(3)/9


the final answer is: sqrt(3)/9




Thanks for writing.


Staff
www.solving-math-problems.com


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