+0  
 
0
47
5
avatar+560 

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

Lightning  Jul 24, 2018
 #1
avatar+560 
+1

Thank you Chris for getting back to me so fast!

Lightning  Jul 24, 2018
 #2
avatar+87564 
+1

5x^2 + 19x + 16  > 20      subtract 20 from both sides

 

5x^2 + 19x - 4   > 0       let's set this to 0

 

5x^2 + 19x - 4  = 0     factor, if we can

 

(5x -1 ) ( x + 4)  = 0      set each factor to 0 and solve for x and we  get

 

x  =1/5     and  x = -4

 

Note that  we have three possible intervals that make the original inequality  true

 

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

 

Normally, in a quadratic...if the middle interval  makes the inequality untrue, the  outside intervals will make it  true

 

Testing a point in the middle interval  - I'll pick  x  = 0 -  and subbing it into the original inequality  produces  16 > 20

 

So....the  middle interval makes the inequality untrue....and the integers in this interval  are  -3, -2, -1 and 0  [ 4 values]

(Note that we  can't include -4 since it is not part of the interval itself)

 

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

 

 

cool cool cool

CPhill  Jul 24, 2018
 #3
avatar+87564 
0

No prob...maybe they should call Me "Lightning,"  huh ??

 

 

cool cool cool

CPhill  Jul 24, 2018
 #4
avatar+560 
0

Lol! Maybe i should! BTW, -4 is included! Thanks!

Lightning  Jul 24, 2018
 #5
avatar+87564 
0

Yep...you are correct...I forgot that we  need to  find the integer values that make the inequality untrue...-4 should be included!!!

 

 

cool cool cool

CPhill  Jul 24, 2018

14 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.