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# ​ Please leave clear solution thx :)

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

Guest Apr 6, 2017
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#1
+80913
+1

This one is a little difficult.....however.......note that x^2 + x + 3  is positive for all x

We can see this because this is a parabola that turns upward and the vertex is  (-1/2, 2.75)  .....which is above the x  axis......so this part of the function is never negative

So.....only the denominator can  make this function negative....and we have

[ 2x^2 + x - 6 ] ≥ 0

Factor the denominator

(2x -3) ( x + 2)  ≥ 0

Setting each factor to  0 and solving for x, we have that   x = 3/2   and x =  -2

And we can see that when    -2 < x < 3/2   the denominator will be  negative

Then...the function will be   ≥ 0  on these intervals :

(-infinity , -2) U (3/2, +infinity)

Notice that we have to exclude  -2 and 3/2  from the soution because these would make the denominator 0

See the graph here :   https://www.desmos.com/calculator/dztw4t5o0m

CPhill  Apr 6, 2017
#2
+1364
+1

Very nice Cphill!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

tertre  Apr 7, 2017

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