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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.

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   
Associated Topics:
High School Basic Algebra

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