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laurad3

math problem

could someone help me out? I haven't worked with absolute values before

3|x| / x-1 < 2

thanks

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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)

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3|x| / x-1 < 2

So first thing we want to do is square both sides.

9x2 /(x2 -2x+1) < 4

Since we know the squared value on the bottom must be positive, we can multiply it through

9x2 < 4x2 -8x+4

Algebraically:

5x2 + 8x -4 < 0

Turn into equation and find roots to find critical numbers.

(x+2)(5x-2)=0

Therefore roots are -2, 2/5

Set up an intreval line with your roots.

-inf............-2.........2/5........inf

0

Test the central interval in your original inequality. Since 0 falls in, use that.

3|x| / x-1 < 2

3|0| / -1 < 2

0 < 2

True. 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.

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