Question 138674
{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality


{{{2x^2-5x-3=0}}} Set the denominator equal to zero



{{{(x-3)(2x+1)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-3=0}}} or  {{{2x+1=0}}} 


{{{x=3}}} or  {{{x=-1/2}}}    Now solve for x in each case



So the vertical asymptotes are
 

{{{x=3}}} or  {{{x=-1/2}}} 




These are the first two critical values



{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality




{{{(x+5)/(2x^2-5x-3)=0}}} Set the left side equal to zero




{{{x=-5}}} Solve for x



So our critical values are {{{x=3}}}, {{{x=-1/2}}} and {{{x=-5}}} 






Now set up a number line and plot the critical values on the number line


{{{number_line( 600, -10, 10,-5,-1/2,3)}}}




So let's pick some test points that are near the critical values and evaluate them.



Let's pick a test value that is less than {{{-5}}} (notice how it's to the left of the leftmost endpoint):


So let's pick {{{x=-6}}}


{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality



{{{(-6+5)/(2(-6)^2-5(-6)-3)>= 0}}} Plug in {{{x=-6}}}



{{{(-1)/99>= 0}}} Evaluate and simplify the left side


Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.



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Let's pick a test value that is in between {{{-5}}} and {{{-1/2}}}:


So let's pick {{{x=-2}}}


{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality



{{{(-2+5)/(2(-2)^2+-5(-2)-3)>= 0}}} Plug in {{{x=-2}}}



{{{1/5>= 0}}} Evaluate and simplify the left side


Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.


   So part our solution in interval notation is <font size="8">[</font>*[Tex \LARGE -5,-\frac{1}{2}]<font size="8">)</font>



   



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Let's pick a test value that is in between {{{-1/2}}} and {{{3}}}:


So let's pick {{{x=1}}}


{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality



{{{(1+5)/(2(1)^2-5(1)-3)>= 0}}} Plug in {{{x=1}}}



{{{-1>= 0}}} Evaluate and simplify the left side


Since the inequality is false, this means that the interval does <b>not</b> work. So this interval is <b>not</b> in our solution set and we can ignore it.



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Let's pick a test value that is greater than {{{3}}} (notice how it's to the right of the rightmost endpoint):


So let's pick {{{x=4}}}


{{{(x+5)/(2x^2-5x-3)>=0}}} Start with the given inequality



{{{(4+5)/(2(4)^2-5(4)-3)>= 0}}} Plug in {{{x=4}}}



{{{1>= 0}}} Evaluate and simplify the left side


Since the inequality is true, this means that the interval works. So this tells us that this interval is in our solution set.

   So part our solution in interval notation is <font size="8">(</font>*[Tex \LARGE 3,\infty]<font size="8">)</font>



   



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Summary:


So the solution in interval notation is:



<font size="8">[</font>*[Tex \LARGE -5,-\frac{1}{2}]<font size="8">)</font> *[Tex \LARGE \cup] <font size="8">(</font>*[Tex \LARGE 3,\infty]<font size="8">)</font>