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For how many integers \(b\) does the inequality \(z^2 + bz + 15 < 0\) have no real solutions \(z\)?

 Jul 23, 2019
 #1
avatar+105488 
+3

This is a parabola that turns upward

We can find the x coordinate of the vertex  as  -b/2

 

So....we can solve this for "b"

 

(-b/2)^2  + b(-b/2) + 15  < 0

 

b^2/4  - b^2/2  + 15 < 0     multiply through by 4

 

b^2 - 2b^2 + 60 < 0

 

-b^2 + 60< 0   multiply through by - 1  and reverse the inequality sign

 

b^2 -60 > 0

 

(b + √60) (b - √60) > 0

 

We have three possible solution intervals

 

(-infinity, -√60) or ( -√60, √60)  or ( √60, infinity)

 

Testing z = 0 in the middle interval in the original inequality makes it untrue

 

So.....the number of integers between -√60  and √60  =  -7  and 7 inclusive  =  15 integers 

 

So....15 integer values for b produce no real solutions for z

 

 

cool cool cool 

 Jul 23, 2019

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