Question 74896
{{{2/ (sqrt(6) - sqrt(5))}}}
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I'm going to assume that both the square root terms are in the denominator as shown above.
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You can rationalize the denominator through using the identity:
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{{{(a + b)*(a - b) = a^2 - b^2}}}
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You can multiply out the left side of this identity just to convince yourself that the result
equals the right side.
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The denominator of the problem is of the form (a - b).  Suppose that we multiply the entire
term of the problem by {{{(a + b)/(a + b)}}} which is equivalent to multiplying the term 
in the problem by 1 since the numerator of this multiplier equals the denominator.  
I used "a" and "b" so it might be easier to see what we're trying to do.  Actually, we're 
going to let:
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{{{a = sqrt(6)}}} and {{{b = sqrt(5)}}}.
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So when we multiply the original problem by  {{{(a + b)/(a + b)}}} we're actually going
to multiply it by {{{(sqrt(6)+sqrt(5))/(sqrt(6)+sqrt(5))}}}.
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Let's do it:
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{{{(2/(sqrt(6) - sqrt(5)))*((sqrt(6)+sqrt(5))/(sqrt(6)+sqrt(5)))}}}
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In accordance with the identity above, after multiplying the original denominator
of the problem by {{{(sqrt(6)+sqrt(5))}}} the new, rationalized denominator becomes
the difference between the squares of the two terms that were in the original denominator:
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{{{(sqrt(6)-sqrt(5))*(sqrt(6)+sqrt(5)) = (sqrt(6))^2 - (sqrt(5))^2}}}
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But {{{(sqrt(6))^2 = 6}}} and {{{(sqrt(5))^2 = 5}}} so the new, rationalized denominator
just equals 6 - 5 or 1.
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Now all we have left to do is to work on the numerator.
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The numerator of the original problem is 2, but we multiplied it by {{{(sqrt(6)+sqrt(5))}}}
as part of the process associated with rationalizing the denominator. When we do this we get:
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{{{2*(sqrt(6)+sqrt(5)) = 2*sqrt(6) + 2*sqrt(5)}}}
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That's the answer ... {{{2*sqrt(6)+2*sqrt(5)}}} because the new denominator that it is over
is just 1.
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Hope that after looking through this you'll be familiar with using the identity to rationalize
denominators.  It will be useful to you when you work with complex numbers.
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