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December 17, 2014

December 17, 2014

Posted by **Anonymous** on Tuesday, October 13, 2009 at 5:24pm.

Would the solutions be x=5 and x= -2?

- Algebra -
**Reiny**, Tuesday, October 13, 2009 at 5:38pmno, you forgot about the inequality condition.

You must have had

(x-5)(x+2) ≤ 0

here is a neat way to finish it

draw a line and mark the "critical values" of -2 and +5

this splits the line into 3 segments

1. x < -2

2. x between -2 and 5

3. x > 5

pick any x you want in each of the segments to test . You don't actually have to find the value just the +/- situation

1. let x = -10 , so (-)(-) > 0 , NO

2. let x = 0 , so (-)(+) < 0 YEs

3. let x = 10, then (+)(+) > 0, NO

so every value of x between -2 and 5 inclusive will work..

-2 ≤ x ≤ 5 , x any real number

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