Thread: quadratic inequality

x^2-2x-8>0

I know that the answer is x>4 or x<-2. My question is why the sign is reversed if there is no multiplication or division on either side?

Consider x^2-2x-8 = (x-4)(x+2). This is > 0 if either x-4 and x+2 are both > 0 or both < 0. The first case leads to x>4 and x>-2, which is just x>4; the second to x<4 and x<-2, which is x<-2. The reversal of sign comes from this second case where you consider the possibility that two negative factors multiply to give a positive value.

Here is one way.

Get the critical points of the inequality, or the points when the graph of the inequality crosses the x-axis. You treat the inequality as an equation only---meaning, forget about the < and > signs. Consider only the = sign.
Then, these critical points subdivides the x-axis into intervals of the x-values.
Then, test/check the inequality in these intervals. The solution/solutions of the inequality are the intervals where the inequality is valid/true.

x^2 -2x -8 > 0

x^2 -2x -8 = 0
(x-4)(x+2) = 0
x = 4, or -2

So there are 3 inervals:
(-infinity,-2), (-2,4), and (4,infinity)

Test the x^2 -2x -8 > 0 in the (-infinity,-2) interval.
Say, x = -3.
(-3)^2 -2(-3) -8 >? 0
9 +6 -8 >? 0
7 >? 0
Yes, so, this interval is a solution. -----***

Test the x^2 -2x -8 > 0 in the (-2,4) interval.
Say, x = 0.
(0)^2 -2(0) -8 >? 0
0 -0 -8 >? 0
-8 >? 0
No, so, this interval is not a solution.

Test the x^2 -2x -8 > 0 in the (4,infinity) interval.
Say, x = 7.
(7)^2 -2(7) -8 >? 0
49 -14 -8 >? 0
27 >? 0
Yes, so, this interval is a solution also. -----***

Therefore,
x = (-infinity,-2) ---or x < -2
x = (4, infinity) ----or x > 4

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What about the critical x's themselves, when x = -2, and when x=4?

The inequality does not say "greater than or equal to" zero, hence x cannot equal -2 0r 4.

If the inequality were x^2 -2x -8 >= 0,
then x <= -2 or x >= 4.