SOLUTION: I don't get how do we find the solution set for {{{x^2-x-2<0}}}. Thank You very much :)

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Question 512295: I don't get how do we find the solution set for . Thank You very much :)
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
We want values of x such that x² - x - 2 < 0, or

(x - 2)(x + 1) < 0

First we find the zeros of the function f(x) = x² - x - 2 

We write it in factored form

f(x) = (x - 2)(x + 1)

Then we'll test the intervals which they divide the number line into.

To find the zeros of f(x), we set f(x) = 0

(x - 2)(x + 1) = 0

x - 2 = 0;  x + 1 = 0

    x = 2;      x = -1

Nest we draw a number line and mark those points on it with an
open circle:

-------------o--------o-------------
-5 -4 -3 -2 -1  0  1  2  3  4  5  6

We pick a test value on the infinitely long interval left of
-1.  The easiest value to pick is -2, and substitute it into
the original inequality

  (x - 2)(x + 1) < 0
(-2 - 2)(-2 + 1) < 0
        (-4)(-1) < 0
               4 < 0   

That is false, so we do not shade the region left of -1, so we
still have this:

-------------o--------o-------------
-5 -4 -3 -2 -1  0  1  2  3  4  5  6

Next we pick a test value on the interval between -1 and 2.  The 
easiest value to pick is 0, and substitute it into
the original inequality

(x - 2)(x + 1) < 0
(0 - 2)(0 + 1) < 0
       (-2)(1) < 0
            -2 < 0 

That is true, so we DO shade the interval between -1 and 2, so we
now have this:

-------------o========o-------------
-5 -4 -3 -2 -1  0  1  2  3  4  5  6

We pick a test value on the infinitely long interval right of
2.  The easiest value to pick is 3, and substitute it into
the original inequality

(x - 2)(x + 1) < 0
(3 - 2)(3 + 1) < 0
        (1)(4) < 0
             4 < 0  

That is false, so we do not shade the region right of 2, so we
end up with this:

-------------o========o-------------
-5 -4 -3 -2 -1  0  1  2  3  4  5  6

The interval notation is an appreviation for the above number
line graph, and this interval notation abbreviation is

              (-1, 2)

[Don't confuse this with the 2D point on the xy-plane which has the
same notation.  This is a conflict in mathematical notation.  Sometimes
(-1,2) means a point in the xy-plane and other times it means an interval
on the number line.  We just have to live with this conflict, and just go
by context to discover which it means.)  

Edwin


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