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Find the number of integers $n$ that satisfy $n^2 < 64 - 20n - n^2.$

 Apr 3, 2024
 #1
avatar+129690 
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Simplify as

 

2n^2 + 20n - 64 <  0

 

n^2 + 10n - 32  <  0       set up as an equality

 

n^2  + 10n  =  32          complete the square on  n

 

n^2 + 10n + 25  = 32 + 25

 

( n + 5)^2  =  57      take  both roots

 

n + 5  =  sqrt (57)      and    n + 5 =  -sqrt (57)

 

n = sqrt (57)  - 5   ≈ 2.5             n  =  -5 -sqrt (57) ≈ -12.5

 

These  are roots of the quadratic which opens upward.....every integer between  them will cause the given inequality to be <  0

 

The number of intgers  =  2  - -12 + 1  =  2 + 12 + 1  =  15

 

cool cool cool

 Apr 3, 2024

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