Absolute Value: Two Methods
Date: 07/03/98 at 10:46:37 From: Presley Subject: Absolute value Hello Dr. Math, I have a problem I'm curious about: Solve for x: |x+3| < |x-2| I wanted to square both sides, but I am scared that it is not correct. Can you show me how to solve this equation?
Date: 07/04/98 at 05:17:12 From: Doctor Anthony Subject: Re: Absolute value For this particular problem you could square both sides, but that technique is not always applicable, for example |x-2| + |x+5| - |x-3| > |x+13| So in the example you gave |x+3| < |x-2| squaring both sides x^2 + 6x + 9 < x^2 - 4x + 4 10x < -5 x < -1/2 Alternative method. ------------------- Critical values are -3 and +2, so we consider 3 separate situations: x < -3 Inequality can be written -(x+3) < -(x-2) -x - 3 < -x + 2 - 3 < 2 (always true) So the inequality is satisfied for all x < -3 Second region is -3 < x < 2 Inequality can be written x+3 < -(x-2) x + 3 < - x + 2 2x < -1 x < -1/2 So the inequality is true for all x < -1/2 Third region is x > 2. Inequality can be written x+3 < x-2 3 < -2 (never true) So no values of x satisfy the inequality in the region x > 2 Combining all three regions we see that the inequality is satisfied when: x < -1/2 This is the same result as before, and in this example the first method is clearly to be preferred. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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