SOLUTION: 5 + square root 3 over 7-2 square root 3

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Question 177662This question is from textbook saxon algebra 2
: 5 + square root 3 over 7-2 square root 3 This question is from textbook saxon algebra 2

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(5 + square root 3) / (7-2 square root 3)
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Multiply numerator and denominator by 7+2sqrt(3)
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[(5+sqrt(3))(7+2sqrt(3)] / [(7-2sqrt(3))(7+2sqrt(3))]
= [35 + 17sqrt(3) + 6] / [47 + 12]
= [41 + 17sqrt(3)] / 59
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Cheers,
Stan H.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I think you mean:



The problem is, you don't say what it is that you would like to do with this expression. How about giving us tutors a break by putting forth the little bit of effort it takes to say what it is you want instead of making us guess?

Having said that, I suspect that you need to rationalize the denominator. If that is the case, then you need to multiply your fraction by 1 in the form of the conjugate of the denominator divided by itself. That way, the radical in the denominator is eliminated by virtue of the fact that the product of a binomial expression and its conjugate results in the difference of two squares.

The conjugate of is (you just change the sign in the middle).

Therefore you need to multiply your original fraction by

So:

Apply FOIL to the numerator factors and the difference of two squares to get the denominator.