SOLUTION: Solve the inequality {{{(x+5)/(2x^2-5x-3)>=0}}}

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Question 138674: Solve the inequality %28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality

2x%5E2-5x-3=0 Set the denominator equal to zero


%28x-3%29%282x%2B1%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:
x-3=0 or 2x%2B1=0

x=3 or x=-1%2F2 Now solve for x in each case


So the vertical asymptotes are

x=3 or x=-1%2F2



These are the first two critical values


%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality



%28x%2B5%29%2F%282x%5E2-5x-3%29=0 Set the left side equal to zero



x=-5 Solve for x


So our critical values are x=3, x=-1%2F2 and x=-5





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

number_line%28+600%2C+-10%2C+10%2C-5%2C-1%2F2%2C3%29



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

%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality


%28-6%2B5%29%2F%282%28-6%29%5E2-5%28-6%29-3%29%3E=+0 Plug in x=-6


%28-1%29%2F99%3E=+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not 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%2F2:

So let's pick x=-2

%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality


%28-2%2B5%29%2F%282%28-2%29%5E2%2B-5%28-2%29-3%29%3E=+0 Plug in x=-2


1%2F5%3E=+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 [)





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Let's pick a test value that is in between -1%2F2 and 3:

So let's pick x=1

%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality


%281%2B5%29%2F%282%281%29%5E2-5%281%29-3%29%3E=+0 Plug in x=1


-1%3E=+0 Evaluate and simplify the left side

Since the inequality is false, this means that the interval does not work. So this interval is not 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

%28x%2B5%29%2F%282x%5E2-5x-3%29%3E=0 Start with the given inequality


%284%2B5%29%2F%282%284%29%5E2-5%284%29-3%29%3E=+0 Plug in x=4


1%3E=+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 ()





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Summary:

So the solution in interval notation is:


[) ()