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November 28, 2014

November 28, 2014

Posted by **black_widow** on Saturday, March 23, 2013 at 12:08am.

where /x/ is the modulus of x.

- maths -
**Steve**, Saturday, March 23, 2013 at 12:29amwhy not just use ||, since / is usually used for division?

||(2x-1)(x+3)|-4| < 1

You have four cases to consider:

|(2x-1)(x+3)|-4 positive or negative, and within those,

(2x-1)(x+3) positive or negative

you will quite possibly end up with a solution set comprising 4 intervals

- maths -
**black_widow**, Saturday, March 23, 2013 at 9:15amstill cant solve it.....

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