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For how many integer values of x is 5x^2+19x+16 > 20 not satisfied?

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Guest Apr 22, 2018
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5x^2+19x+16 > 20      subtract 20  from both sides

 

5x^2 + 19x - 4  >  0     factor

 

(5x - 1) ( x + 4)  > 0

 

Setting both factors to 0,.....we can establish 3 possible intervals

 

5x - 1  = 0            x + 4  = 0

5x  = 1                 x = -4

x  =  1/5

 

The possible intervals that provide solutions are

 

(-inf, - 4)  U ( - 4, 1/5)  U ( 1/5, inf )

 

And either the middle interval solves the original inequality or the two outside intervals do

 

Testing a point in the middle interval  - I'll pick 0 -  in the original inequality we have

 

5(0)^2 + 19(0)   + 16  >  20      is false

 

So...since we are looking for the integer that make the inequality false....the solution will be all  the integers  on  ( - 4, 1/5)

 

 

And the integers in this interval  are  -3, -2, -1, 0

 

So.... 4 integers do not satisfy this

 

 

cool cool cool

CPhill  Apr 22, 2018

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