SOLUTION: I need your help on solving an inequality. I know how to solve it but I don't understand the test points. For example: x^2+6x<16 I know that the problem gets put in standard form

Algebra ->  Inequalities -> SOLUTION: I need your help on solving an inequality. I know how to solve it but I don't understand the test points. For example: x^2+6x<16 I know that the problem gets put in standard form      Log On


   

Question 187773: I need your help on solving an inequality. I know how to solve it but I don't understand the test points. For example:
x^2+6x<16
I know that the problem gets put in standard form
x^2+6x-16<0
and then it is factored
(x+8)(x-2)
then you set the factors equal to zero for the x intercepts
x+8=0 x=-8 and x-2=0 x=2
but I don't understand the test points
for example
(-2,8)
(-8,2)
(-8,-2)U (8,8)
(-8,-8)U (2,8)
please explain
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You're off to a great start.

Now construct a number line with the critical values plotted on the line:




Notice how there are three regions:


Region #1: To the left of the first critical value -8. The region in interval notation is ()

Region #2: In between the critical values -8 and 2. The region in interval notation is ()

Region #3: To the right of the second value 2. The region in interval notation is ()



For any region, the graph is either above the x-axis or below the x-axis (no region is a mix of the two or on the x-axis). So all we have to do is plug in test points that represent the three regions to find the solution set.


-----------------------------------------------

Let's see if the first region is part of the solution set.


%28x%2B8%29%28x-2%29%3C0 Start with the factored expression


%28-9%2B8%29%28-9-2%29%3C0 Plug in x=-9 (this value is less than -8 which means that it lies in the first region)


%28-1%29%28-11%29%3C0 Combine like terms.


11%3C0 Multiply


Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy x%5E2%2B6x%3C16. So we can ignore this region.


------------------------------------------------------------


Let's see if the second region is part of the solution set.


%28x%2B8%29%28x-2%29%3C0 Start with the factored expression


%28-7%2B8%29%28-7-2%29%3C0 Plug in x=-7 (this value is greater than -8 and less than 2)


%281%29%28-9%29%3C0 Combine like terms.


-9%3C0 Multiply


Since this inequality is TRUE, this means that ANY x value in this region satisfies x%5E2%2B6x%3C16. So the interval () is part of the solution set.


----------------------------------------------------------------


Let's see if the third region is part of the solution set.


%28x%2B8%29%28x-2%29%3C0 Start with the factored expression


%283%2B8%29%283-2%29%3C0 Plug in x=3 (this value is greater than 2)


%2811%29%281%29%3C0 Combine like terms.


11%3C0 Multiply


Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy x%5E2%2B6x%3C16. So we can ignore this region.


===================================================================


Answer:


So the solution set to the inequality x%5E2%2B6x%3C16 is ()


If the above did not make any sense, then take a look at the graph of y=x%5E2%2B6x-16




Notice that the x-intercepts are -8 and 2. The portion of the graph to the left of -8 is ALL above the x-axis. The portion in between -8 and 2 is ALL below the x-axis. Finally, the piece from 2 on to infinity is ALL above the x-axis.


So the only portion that satisfies x%5E2%2B6x-16%3C0 is from -8 to 2.