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For how many integer values of x is \(5x^{2}+19x+16 > 20\) not satisfied?

 Jul 26, 2019
 #1
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+1

 

If you start with what you're given    5x2 + 19x + 16  > 20

 

Subtract 20 from both sides             5x2 + 19x – 4  > 0 

 

Factor the left side                           (5x – 1)(x + 4)  > 0 

 

I thought I I'd figure out the answer as I went,

but I find that I don't know what to do now.

.

 Jul 26, 2019
 #2
avatar+36916 
0

Building on Guest's answer  (Thanx!)

 

Between x = -4 and + 1/5   the graph is neg   or zero  

 

Below -4   or above +1/5 it is > 0

 

so    (-inf,-4) U (1/5, + inf)    it is greater than zero  (or the original equation is > 20)

 

EDIT:   So NOT satisfied would be     [-4,1/5]    -4, -3,-2,-1 0    FIVE integer values

 Jul 26, 2019
edited by ElectricPavlov  Jul 26, 2019
edited by ElectricPavlov  Jul 26, 2019
edited by ElectricPavlov  Jul 26, 2019

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