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Unformatted text preview: rk a couple more examples that are a little more difficult. Example 2 Solve each of the following equations. (a) y + y - 4 = 4 [Solution] (b) 1 = t + 2t - 3 [Solution] (c) 5 z + 6 - 2 = z [Solution] Solution (a) y + y - 4 = 4 In this case let’s notice that if we just square both sides we’re going to have problems. (y+ y-4 ) 2 = ( 4) 2 y 2 + 2 y y - 4 + y - 4 = 16 Before discussing the problem we’ve got here let’s make sure you can do the squaring that we did above since it will show up on occasion. All that we did here was use the formula (a + b) 2 = a 2 + 2ab + b 2 with a = y and b = y - 4 . You will need to be able to do these because while this may not have worked here we will need to this kind of work in the next set of problems. Now, just what is the problem with this? Well recall that the point behind squaring both sides in the first problem was to eliminate the square root. We haven’t done that. There is still a square root in the problem and we’ve make the remainder of the problem messier as well. © 2007 Paul Dawkins 117 http://tutor...
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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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