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For what values of $x$ is  $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$

Guest Feb 28, 2018
 #1
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\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + (4-3)x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + 4x-3x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x( x+2) - 3(x+2)} \ge 0\\ \frac{x^2 + x + 3}{(2x-3)( x+2) } \ge 0\\ \)

 

The denomator cannot be equal to zero so x cannot equal  3/2   or  -2

 

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The denominator by itself is  positive  for    \(-2

and it is negative for  \(x<-2\) and   negative for  \(x>1.5\)

 

Now I will look at just the numerator.  

consider \(y=x^2+x+3\)

 

this is a concave up parabola

 

The zeros will be  

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-1 \pm \sqrt{1-12} \over 2}\\ \text{There are no real zeros} \)

 

this means that th numerator is positive for all real values of x.

 

Since the numerator is positive the quotient will be positive when and only when the denominator is positive.

 

So this expression will be greater than zero when     \(-2

Melody  Feb 28, 2018
 #2
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What is \(-2?

Guest Feb 28, 2018

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