Mathlamaredutermsaspx college algebra to this point in

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Unformatted text preview: the integer and/or fractional solutions. In this case they are, t= -19 + 23 2 = 14 7 t= -19 - 23 = -3 14 Now, as with completing the square, the fact that we got integer and/or fractional solutions means that we could have factored this quadratic equation as well. [Return to Problems] (d) 3 1 = +1 y-2 y So, an equation with fractions in it. The first step then is to identify the LCD. LCD : y ( y - 2 ) So, it looks like we’ll need to make sure that neither y = 0 or y = 2 is in our answers so that we don’t get division by zero. Multiply both sides by the LCD and then put the result in standard form. æ 3 ö æ1 ö ÷ = ç + 1÷ ( y )( y - 2 ) è y-2ø è y ø ( y )( y - 2 ) ç 3 y = y - 2 + y ( y - 2) 3y = y - 2 + y2 - 2 y 0 = y2 - 4 y - 2 Okay, it looks like we’ve got the following values for the quadratic formula. a =1 b = -4 c = -2 Plugging into the quadratic formula gives, © 2007 Paul Dawkins 100 http://tutorial.math.lamar.edu/terms.aspx College Algebra y= - ( -4 ) ± ( -4 ) - 4 (1)( -2 ) 2 (1) 2 4 ± 24 2 4±2 6 = 2 = 2± 6 = Note that both of these are going to be solutions since neither of them are the values that we need to avoid. [Return to Problems] (e) 16 x - x 2 = 0 We saw an equation similar to this in the previous section when we were looking at factoring equations and it would definitely be ea...
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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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