Alg_Complete

# Mathlamaredutermsaspx college algebra absolute value

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Unformatted text preview: here the inequality is zero, so this means we include x = -1 . Notice that we will also need to avoid x = 5 since that gives division by zero. The solution for this inequality is, [ -1,5) -1 £ x < 5 Example 2 Solve x2 + 4 x + 3 >0. x -1 Solution We’ve got zero on one side so let’s first factor the numerator and determine where the numerator and denominator are both zero. ( x + 1) ( x + 3) > 0 numerator : x = -1, x = -3 x -1 denominator : x = 1 Here is the number line for this one. In the problem we are after values of x that make the inequality strictly positive and so that looks like the second and fourth region and we won’t include any of the endpoint here. The solution is then, -3 < x < -1 1< x < ¥ ( -3, -1) © 2007 Paul Dawkins and and (1, ¥ ) 136 http://tutorial.math.lamar.edu/terms.aspx College Algebra Example 3 Solve x 2 - 16 ( x - 1) 2 < 0. Solution There really isn’t too much to this example. We’ll first need to factor the numerator and then determine where the numerator and denominator are zero. ( x - 4) ( x + 4) < 0 2 ( x - 1) numerator : x = -4, x = 4 denomin...
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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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