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I can't solve this

 

For an integer n, the inequality x^2 + nx + 15 < -21 - 280
has no real solutions in x. Find the number of different possible values of n.

 Jun 26, 2024
 #1
avatar+129739 
+1

Simplify as

 

x^2 + nx + 316 < 0

 

Set as an equality 

 

We will have real solutions whenever the discriminant is  ≥ 0

 

So

 

n^2 - 4(1)(316) ≥ 0

 

n^2 - 1264 ≥ 0

 

n^2 ≥ 1264             take both roots

 

n ≥ sqrt (1264)  → ≈ 35.6   or    n  ≤  -sqrt (1264) → ≈  -35.6

 

Thus....we have real solutions when  n  = ( inf , ≈ -35.6] U [ ≈ 35.6 , inf )

 

So  the integers for n that  make this  have NO real solutions  in x are the integers between these two values

n = [ -35 , 35 ]  =    2(35) + 1 =   71  integers

 

cool cool cool

 Jun 26, 2024

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