So provided we can factor a polynomial we can always

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Unformatted text preview: (a) 1 5 = 1x +1 2x - 4 The LCD for this problem is ( x + 1) ( 2 x - 4 ) and we will need to avoid x = -1 and x = 2 to make sure we don’t get division by zero. Here is the work for this equation. ( x + 1)( 2 x - 4 ) æ ç 1ö 5ö æ ÷ = ( x + 1)( 2 x - 4 ) ç1 ÷ è x +1ø è 2x - 4 ø 2 x - 4 = ( x + 1)( 2 x - 4 ) - 5 ( x + 1) 2 x - 4 = 2 x2 - 2 x - 4 - 5x - 5 0 = 2 x2 - 9 x - 5 0 = ( 2 x + 1)( x - 5 ) So, it looks like the two solutions to this equation are, x=- 1 2 and x=5 Notice as well that neither of these are the values of x that we needed to avoid and so both are solutions. [Return to Problems] 3 4- x = x -1 x -1 In this case the LCD is x - 1 and we will need to avoid x = 1 so we don’t get division by zero. (b) x + 3 + Here is the work for this problem. ( x - 1) æ x + 3 + ç 3 ö æ 4- xö ÷=ç ÷ ( x - 1) x -1 ø è x -1 ø è ( x - 1)( x + 3) + 3 = 4 - x x2 + 2 x - 3 + 3 = 4 - x x 2 + 3x - 4 = 0 ( x - 1)( x + 4 ) = 0 So, the quadratic that we factored and solved has two solutions, x = 1 and x = -4 . However, when we...
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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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