Question 187773
You're off to a great start. 


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


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/number_line.png">



Notice how there are three regions: 



Region #1: To the left of the first critical value -8. The region in interval notation is <font size="6">(</font>*[Tex \LARGE -\infty,-8]<font size="6">)</font>


Region #2: In between the critical values -8 and 2. The region in interval notation is <font size="6">(</font>*[Tex \LARGE -8,2]<font size="6">)</font>


Region #3: To the right of the second value 2. The region in interval notation is <font size="6">(</font>*[Tex \LARGE 2,\infty]<font size="6">)</font>




For <i>any</i> 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.



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



{{{(x+8)(x-2)<0}}} Start with the factored expression



{{{(-9+8)(-9-2)<0}}} Plug in {{{x=-9}}} (this value is less than -8 which means that it lies in the first region)



{{{(-1)(-11)<0}}} Combine like terms.



{{{11<0}}} Multiply



Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy {{{x^2+6x<16}}}. So we can ignore this region.



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



{{{(x+8)(x-2)<0}}} Start with the factored expression



{{{(-7+8)(-7-2)<0}}} Plug in {{{x=-7}}} (this value is greater than -8 and less than 2)



{{{(1)(-9)<0}}} Combine like terms.



{{{-9<0}}} Multiply



Since this inequality is TRUE, this means that ANY x value in this region satisfies {{{x^2+6x<16}}}. So the interval <font size="6">(</font>*[Tex \LARGE -8,2]<font size="6">)</font> is part of the solution set.



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



{{{(x+8)(x-2)<0}}} Start with the factored expression



{{{(3+8)(3-2)<0}}} Plug in {{{x=3}}} (this value is greater than 2)



{{{(11)(1)<0}}} Combine like terms.



{{{11<0}}} Multiply



Since this inequality is FALSE, this means that ANY x value in this region does NOT satisfy {{{x^2+6x<16}}}. So we can ignore this region.



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Answer: 



So the solution set to the inequality {{{x^2+6x<16}}} is <font size="6">(</font>*[Tex \LARGE -8,2]<font size="6">)</font>



If the above did not make any sense, then take a look at the graph of {{{y=x^2+6x-16}}}


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/graph2.png">



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^2+6x-16<0}}} is from -8 to 2.