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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.
 Spring '12
 MrVinh

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