+1206 For how many integers \(b\) does the inequality \(z^2 + bz + 15 < 0\) have no real solutions \(z\)?
+128460 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
