Question 167075: how do you solve this problem and what is the answer : square root of 6 over squar root of 5 - squar root of 3
Answer by midwood_trail(310)   (Show Source):
You can put this solution on YOUR website! Let sqrt = square root
sqrt{6}/(sqrt{5} - sqrt{3})
Multiply the top and bottom by [sqrt{5} + sqrt{3}]
sqrt{6} times sqrt{5} + sqrt{3} = sqrt{30} + sqrt{18}...Our numerator
[sqrt{5} - sqrt{3}] times [sqrt{5} + sqrt{3}] = sqrt{25} - sqrt{9}...our denominator
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We now have the following fraction:
[sqrt{30} + sqrt{18}]/[sqrt{25} - sqrt{9}]
In the denominator, we have two perfect squares.
So, sqrt{25} = 5 and sqrt{9} = 3
Then 5 - 3 = 2.
In the numerator, sqrt{30} is already in lowest terms and so it stays the same.
However, sqrt{18} becomes 3(sqrt{2}).
Final answer: [3(sqrt{2}) + sqrt{30}]/2
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