laurad3 Posted March 19, 2013 Report Share Posted March 19, 2013 could someone help me out? I haven't worked with absolute values before3|x| / x-1 < 2thanks Reply Link to post Share on other sites More sharing options...
ChocolateDrop Posted March 19, 2013 Report Share Posted March 19, 2013 (edited) Hint: Whenever you have modulus you start out by squaring the whole expression. Then you can treat it how you would treat a normal one. (this one goes through the squaring method) Edited March 19, 2013 by ChocolateDrop Reply Link to post Share on other sites More sharing options...
Rahul Posted March 19, 2013 Report Share Posted March 19, 2013 3|x| / x-1 < 2So first thing we want to do is square both sides.9x2 /(x2 -2x+1) < 4Since we know the squared value on the bottom must be positive, we can multiply it through9x2 < 4x2 -8x+4Algebraically:5x2 + 8x -4 < 0Turn into equation and find roots to find critical numbers.(x+2)(5x-2)=0Therefore roots are -2, 2/5Set up an intreval line with your roots.-inf............-2.........2/5........inf 0Test the central interval in your original inequality. Since 0 falls in, use that.3|x| / x-1 < 23|0| / -1 < 20 < 2True. So the central zone is true and the other zones are false.That means the solution to your inequality is the interval (-2,2/5). This interval is open because of the < sign being open itself in the original inequality. Reply Link to post Share on other sites More sharing options...
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