SOLUTION: I worked out the inequality and found the possible solutions but do not feel like I am writing the interval notation correctly. I would appreciate your help and a big thank you in

Algebra ->  Inequalities -> SOLUTION: I worked out the inequality and found the possible solutions but do not feel like I am writing the interval notation correctly. I would appreciate your help and a big thank you in      Log On


   

Question 148164: I worked out the inequality and found the possible solutions but do not feel like I am writing the interval notation correctly. I would appreciate your help and a big thank you in advance. 6x^2+25x+14<=0
In the end I have -2/3 and -7/2 so far I have ( -7/2] U [-2/3) with the infinities before the -7/2 and the -2/3.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!


6x%5E2%2B25x%2B14%3C=0 Start with the given inequality



%282x%2B7%29%283x%2B2%29%3C=0 Factor the left side


%282x%2B7%29%283x%2B2%29=0 Set the left side equal to zero


Set each individual factor equal to zero:

2x%2B7=0 or 3x%2B2=0

Solve for x in each case:

x=-7%2F2 or x=-2%2F3


So our critical values are x=-7%2F2 and x=-2%2F3

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

number_line%28+600%2C+-10%2C+10%2C-7%2F2%2C-2%2F3%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 -7%2F2 (notice how it's to the left of the leftmost endpoint):

So let's pick x=-4

6x%5E2%2B25x%2B14%3C=0 Start with the given inequality


%282%28-4%29%2B7%29%283%28-4%29%2B2%29%3C=+0 Plug in x=-4


10%3C=+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 -7%2F2 and -2%2F3:

So let's pick x=-2

6x%5E2%2B25x%2B14%3C=0 Start with the given inequality


%282%28-2%29%2B7%29%283%28-2%29%2B2%29%3C=+0 Plug in x=-2


-12%3C=+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 greater than -2%2F3 (notice how it's to the right of the rightmost endpoint):

So let's pick x=1

6x%5E2%2B25x%2B14%3C=0 Start with the given inequality


%282%281%29%2B7%29%283%281%29%2B2%29%3C=+0 Plug in x=1


45%3C=+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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Summary:

So the solution in interval notation is:


[]





Here's a graph to prove it
+graph%28+500%2C+500%2C+-10%2C+10%2C+-10%2C+10%2C+%282x%2B7%29%283x%2B2%29%29+