Once we figure this out we will multiply the

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Unformatted text preview: on] x 1 (c) [Solution] 3- x 5 (d) [Solution] 4 x+ 3 Solution There are really two different types of problems that we’ll be seeing here. The first two parts illustrate the first type of problem and the final two parts illustrate the second type of problem. © 2007 Paul Dawkins 22 http://tutorial.math.lamar.edu/terms.aspx College Algebra Both types are worked differently. 4 x (a) In this case we are going to make use of the fact that n a n = a . We need to determine what to multiply the denominator by so that this will show up in the denominator. Once we figure this out we will multiply the numerator and denominator by this term. Here is the work for this part. x 4x 4x = = x x x2 4 4 = x x Remember that if we multiply the denominator by a term we must also multiply the numerator by the same term. In this way we are really multiplying the term by 1 (since a = 1 ) and so aren’t a changing its value in any way. [Return to Problems] (b) 5 2 x3 We’ll need to start this one off with first using the third property of radicals to eliminate the fraction from underneath the radical as is required for simplification. 5 2 = x3 5 5 2 x3 Now, in order to get rid of the radical in...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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