Setting up Equations
Date: 03/10/2002 at 04:54:53 From: Gideon Subject: Setting up equations correctly Dear Dr Math, I have two questions I am having trouble getting started with: 1) If X^2 > X what can you say about the value of X? What I don't understand is how to set up the equation with a greater than sign (as there is no = sign) and then how to factorise it. I assume that this is a quadratic so there will be two solutions. 2) If X^2 < |X|, what can you say about the value of X? Again I don't understand how to set up the equation and what to do with the absolute. How do I test the equation when there is an absolute value? Any help would be much appreciated. Regards, Gideon
Date: 03/12/2002 at 10:16:56 From: Doctor Douglas Subject: Re: Setting up equations correctly Hi, Gideon. Thanks for submitting your question to the Math Forum. For (1): You've already written the mathematical relation: X^2 > X. To test the values of X at which the inequality possibly changes from a true statement to a false statement, or vice versa, we simply make it an equation: X^2 = X. Solving this equation won't give you the final answer, but it will tell you where you need to check various values for X. This equality has two solutions: X = 0 and X = 1. So from this analysis so far, there are three regions to check on the number line: X<0, 0<X<1, and X>1. Let's test the first region: we choose X = -3 < 0, and see that X^2 = 9 > -3 = X. So in this first region, we have values of X that do satisfy the inequality. You now have to check the other two regions to obtain all of the valid regions for X. For (2): Again, you've already written the mathematical relation: X^2 < |X|. Now the trick is to convert the absolute value sign into expressions that we can work with. Of course, we return to the original definitions when we have to: |X| = X when X >= 0, and |X| = -X when X < 0. So we can write X^2 < X whenever X >= 0 OR X^2 < -X whenever X < 0. All potential values for X will fall in one of these two cases (X either greater than or equal to zero, OR less than zero). For the first of these cases (X^2 < X) you've already done similar work above in problem (1). That case was for X^2 > X, though, so we need to adapt your analysis for this problem. The zero crossings of the equation are the same (X = 0 and X = 1) and we test X = 1/2 to see that X^2 = 1/4 is indeed less than 1/2 = X, and so it is the region 0 < X < 1 that satisfies the first case. Now you need to treat the case X < 0 in a similar fashion to find any remaining possibilities for values of X that satisfy the original inequality with absolute values. Don't forget to check your results by testing a few values for X when you are done! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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