+0  
 
0
113
1
avatar

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

off-topic
Guest Apr 22, 2018
 #1
avatar+91146 
+1

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

12 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.