```Question 74086
If you are given two numbers x and y, the geometric mean between those two numbers is defined
as {{{+sqrt(x*y)}}} and {{{-sqrt(x*y)}}}.
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This means that you just multiply the two given numbers together, take the square root of
the product, simplify the answer, and precede it by a plus-minus sign or write it as
2 identical answers except that one has a plus sign and the other has a minus sign.
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In this problem you are given the two numbers {{{2*sqrt(2)}}} and {{{6*sqrt(6)}}}.
If you
multiply these two together you get:
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{{{2*sqrt(2)*6*sqrt(6)}}}
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The 2 and the 6 multiply to give 12 and the multiplication {{{sqrt(2)*sqrt(6)}}} equals {{{sqrt(12)}}}
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At this stage the answer is {{{12*sqrt(12)}}}. But the radical term can be simplified as
follows:
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{{{sqrt(12) = sqrt(4*3) = sqrt(4)*sqrt(3) = 2*sqrt(3)}}}. Substituting this for {{{sqrt(12)}}} leads
to the product being:
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{{{12*2*sqrt(3)}}} which further simplifies to {{{24*sqrt(3)}}}
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This is now the product that you take the square root of. So you now need to find:
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{{{sqrt(24*sqrt(3))}}}
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This can be split into
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{{{sqrt(24) * sqrt(sqrt(3))}}}
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One factor at a time. First {{{sqrt(24)}}}
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{{{sqrt(24) = sqrt(4*6) = sqrt(4)*sqrt(6) = 2*sqrt(6) = 2*sqrt(2*3) = 2*sqrt(2)*sqrt(3)}}}
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Now for the second term {{{sqrt(sqrt(3))}}}
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Note that taking the square root of the square root of three is the same as:
{{{(3^(1/2))^(1/2)}}} which becomes {{{(3^(1/4))}}}.
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So now our geometric mean (both factors together) has become:
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{{{2*sqrt(2)*sqrt(3)*(3^(1/4))}}}
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Change the {{{sqrt(3)}}} to exponential form: {{{sqrt(3) = 3^(1/2)}}}. Multiply this by the
other exponential form of 3 (remember multiplying means adding exponents) to get:
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{{{2*sqrt(2)*3^(1/2)*3^(1/4) = 2*sqrt(2) * 3^(3/4)}}}
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Finally, {{{2^1 *2^(1/2) = 2^(3/2)}}}
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Substitute this and the geometric mean becomes:
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{{{(2^(3/2))*(3^(3/4))}}} <--- if the numerators in the exponents are cut off by the print generator,
be aware that the exponents are (3/2) and (3/4).
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