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\( {x(x+1) \over 4x^{2}+6x+3}\) it's not factorable, so i'm a little lost!

Guest Aug 9, 2018
 #1
avatar+93644 
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\({x(x+1) \over 4x^{2}+6x+3}   \)

 

You cannot divide by 0 so the denominator cannot be 0

 

\(x \ne {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x \ne {-6 \pm \sqrt{36-48} \over 8}\\ x \ne {-6 \pm \sqrt{12} \;i\over 8}\\ \)

 

This is a complex number so there are no restriction on x in the real number system.

 

There answer has restricted values but I do not think that this is what the question is asking.

Here is the graph if the expression was put equal to y.

You can see that the range is quite restricted.   The domain is not restricted.

 

Melody  Aug 9, 2018

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