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For what values of  is

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?\)
Note: Be thorough and explain why all points in your answer are solutions and why all points outside your answer are not solutions.

 

I looked at Melody's answer but I still don't get it.

 Jun 29, 2018
 #1
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 x^2 + x + 3 

___________    ≥   0

2x^2  + x - 6

 

Look at the graph of   x^2 + x + 3  here :

https://www.desmos.com/calculator/grdcjwwkrt

Note that it is positive for all x...so...the only thing that can make this function negative are x values that make the denominator negative   because 

 

 (positive) 

________   =   negative

(negative)

 

So  factor the bottom  function  and set it to  0   and we have

 

(2x - 3) ( x + 2)   = 0

 

Set  each factor to  0 and solve

 

2x - 3  = 0           x + 2  = 0

2x  =3                 x  = - 2

x  = 3/2 

 

So...we  have the  following values on the number line    x  =- 2  and  x  =3/2

 

And  the  values that  make  2x^2 + x - 6   negative  will come from these possible intervals :

 

(-infinity, -2)   or  ( -2, 3/2)  or  ( 3/2, infinity)

Let's pick a test point in the middle interval, say, 0  and put it into the function

2(0)^2 + 0  -  6   =  -2.....so...in the interval  ( -2, 3/2)....the  bottom  function is negative

You can check and  see that the other two intervals make the function positive

 

So.....the  x values that  make  the  given  function ≥ 0   are

 

(-infinity, -2)   U  ( 3/2, infinity)

 

See the graph that confirms this, here :

 

https://www.desmos.com/calculator/ckhao5mwig

 

 

cool cool cool

 Jun 29, 2018

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