SOLUTION: What is/are the value/s of the inequality {{{(2x^2+6x-3)/(x^2+x-13)>=1}}} in interval notation? Thank you!

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Question 283152: What is/are the value/s of the inequality in interval notation?
Thank you!

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!


Subtract 1 from both sides to get 0 on the right



Write  as 



Multiply top and bottom of  by LCD 





Combine the two numerators over the common denominator:



Remove the parentheses:



Combine like terms on top:



Find all critical values by setting numerator and
denominator = 0.

Setting numerator = 0



discriminant = 

discriminant = 

discriminant is negative, so we don't
get any critical values from the numerator.

Setting denominator = 0



discriminant = 

discriminant = 

So we do get critical values from the
denominator:









So the two critical values are

 and 

These are approximately 3.14 and -4.14

Put these on a number line:

------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5  

Choose a whole number value left of -4.14.
for a test value.  The easiest is -5

Substitute in 






That is a positive number, so we put + signs over 
that part of the number line:

 + + +
------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Next choose a whole number value between the
two critical points for a test value.  The 
easiest is 0

Substitute in 






That is negative so we put - signs over that part 
of the number line:

 + + +  - - - - - - - - - - 
------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Choose a whole number value right of 3.14.
for a test value.  The easiest is 4

Substitute in 





That is a positive number, so we put + signs over 
that part of the number line:

 + + +  - - - - - - - - - -    + + +
------o----------------------o------
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Since the inequality of 



is 

we choose the intervals with + signs.

We cannot include the end points themselves

because they cause the denominator to be 0.

So the solution set is

(,) U (, )

Edwin


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