Rationalizing the DenominatorDate: 07/10/2003 at 18:52:04 From: Indy Bowe Subject: Algebra Question: 1 divided by (the square root of 3 plus the square root of 5 plus the square root of 7). The answer the book gives is: 9 times the sq root of 3, plus 5 times the sq root of 5, plus the sq root of 7, minus 2 times the sq root of 105. Date: 07/10/2003 at 21:35:17 From: Doctor Achilles Subject: Re: Algebra Hi, Thanks for writing to Dr. Math. The problem that you are doing isn't one about adding square roots, it's about a process called "rationalizing the denominator." One thing that comes in handy is knowing how to multiply square roots. When you multiply square roots, just multiply the numbers inside. For example: sqrt(5) * sqrt(3) = sqrt(15) Let me give you an easier example of rationalizing the denominator first that only involves two square roots in the denominator instead of three: 1 ------------------- sqrt(3) + sqrt(5) To get rid of all the sqrts in the denominator, let's multiply by: sqrt(3) - sqrt(5) ------------------- sqrt(3) - sqrt(5) We can do this because we are really just multiplying by 1 since the numerator and denominator are the same. So we have: 1 sqrt(3) - sqrt(5) ------------------- * ------------------- sqrt(3) + sqrt(5) sqrt(3) - sqrt(5) The numerator just becomes: sqrt(3) - sqrt(5) For the denomoniator, we have to multiply each term in the first part by each term in the second part: sqrt(3) * sqrt(3) equals 3 sqrt(3) * -sqrt(5) equals -sqrt(15) sqrt(5) * sqrt(3) equals sqrt(15) sqrt(5) * -sqrt(5) equals -5 So our denominator is: 3 - sqrt(15) + sqrt(15) - 5 Which simplifies to: 3 - 5 or -2 So our fraction is: sqrt(3) - sqrt(5) ------------------- -2 We can multiply the fraction by: -1 ---- -1 Which will give us: -sqrt(3) + sqrt(5) -------------------- 2 Or: sqrt(5) - sqrt(3) ------------------- 2 Now let's use that method to answer your question, let's move on to the tougher one: 1 ----------------------------- sqrt(3) + sqrt(5) + sqrt(7) To answer that, let's multiply by: sqrt(3) - sqrt(5) - sqrt(7) ----------------------------- sqrt(3) - sqrt(5) - sqrt(7) The numerator is: sqrt(3) - sqrt(5) - sqrt(7) For the denominator, we have to multiply each term in the first part by each term in the second part. sqrt(3) * sqrt(3) = 3 sqrt(3) * -sqrt(5) = -sqrt(15) sqrt(3) * -sqrt(7) = -sqrt(21) sqrt(5) * sqrt(3) = sqrt(15) sqrt(5) * -sqrt(5) = -5 sqrt(5) * -sqrt(7) = -sqrt(35) sqrt(7) * sqrt(3) = sqrt(21) sqrt(7) * -sqrt(5) = -sqrt(35) sqrt(7) * -sqrt(7) = -7 The -sqrt(15) and the sqrt(15) will cancel and so will the -sqrt(21) and the sqrt(21), so our denominator will be: 3 - 5 - sqrt(35) - sqrt(35) - 7 Which will simplify to: -9 - 2*sqrt(35) So now we have: sqrt(3) - sqrt(5) - sqrt(7) ----------------------------- -9 - 2*sqrt(35) So now we have a square root in the denominator. So what do we do with it? We have to rationalize (again). So multiply by: -9 + 2*sqrt(35) ----------------- -9 + 2*sqrt(35) Which when all is said and done should give you: -9sqrt(3) + 9sqrt(5) + 9sqrt(7) + 2sqrt(105) - 10sqrt(7) - 14sqrt(5) ---------------------------------------------------------------------- -59 [Please write back and let me know if you get something different.] This should all simplify to: -9*sqrt(3) - 5*sqrt(5) - sqrt(7) + 2*sqrt(105) ------------------------------------------------ -59 Which you can multiply by: -1 ---- -1 To get: 9*sqrt(3) + 5*sqrt(5) + sqrt(7) - 2*sqrt(105) ------------------------------------------------ 59 - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/ |
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