1 1 3 x 3 x 2007 paul dawkins 3 x 3 x 3

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Unformatted text preview: the denominator we need the exponent on the x to be a 5. 5 This means that we need to multiply by 5 2 = x3 5 5 x 2 so let’s do that. 2 5 x2 x3 5 x2 = 5 5 2 x2 x5 = 5 2 x2 x [Return to Problems] (c) 1 3- x In this case we can’t do the same thing that we did in the previous two parts. To do this one we will need to instead to make use of the fact that ( a + b ) ( a - b ) = a2 - b2 When the denominator consists of two terms with at least one of the terms involving a radical we will do the following to get rid of the radical. 1 1 = 3- x 3- x ( © 2007 Paul Dawkins 3+ x = 3+ x ) (3 + x ) (3 - x ) (3 + x ) 23 = 3+ x 9- x http://tutorial.math.lamar.edu/terms.aspx College Algebra So, we took the original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term. By doing this we were able to eliminate the radical in the denominator when we then multiplied out. [Return to Problems] (d) 5 4 x+ 3 This one works exactly the same as the previous example. The only difference is that both terms in the denominator now have radicals...
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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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