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
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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) (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.
Start with the factored expression
Plug in (this value is less than -8 which means that it lies in the first region)
Combine like terms.
Multiply
Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy . So we can ignore this region.
------------------------------------------------------------
Let's see if the second region is part of the solution set.
Start with the factored expression
Plug in (this value is greater than -8 and less than 2)
Combine like terms.
Multiply
Since this inequality is TRUE, this means that ANY x value in this region satisfies . So the interval () is part of the solution set.
----------------------------------------------------------------
Let's see if the third region is part of the solution set.
Start with the factored expression
Plug in (this value is greater than 2)
Combine like terms.
Multiply
Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy . So we can ignore this region.
===================================================================
Answer:
So the solution set to the inequality is ()
If the above did not make any sense, then take a look at the graph of
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 is from -8 to 2.
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