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Find the number of integers \(n\) that satisfy \(7 \sqrt{-n^2 + 22n - 21} \le n + 39.\)

 Apr 15, 2019
 #1
avatar+107424 
+2

 

 

Find the number of integers n that satisfy    \(7 \sqrt{-n^2 + 22n - 21} \le n + 39.\)

 

 

Mmm

firstly:

 

\(-n^2+22n-21\ge0\\ n^2-22n+21\le0\\ (n-21)(n-1)\le0\\ 1\le n\le 21\)

 

now

\(7 \sqrt{-n^2 + 22n - 21} \le n + 39\\ 49(-n^2 + 22n - 21) \le n^2+78n + 1521\\ -50n^2 +49*22n-78n - 49*21-1521 \le 0\\ -50n^2 +1000n - 2550 \le 0\\ 50n^2 -1000n + 2550 \ge 0\\ n^2 -20n + 51 \ge 0\\ (n-17)(n-3) \ge 0\\ 0\le n \le 3 \qquad or \qquad n\ge17\)

 

Take the intersection of these restrictions and I get

n can equal    1, 2, 3, 17, 18, 19, 20 or  21

 

You need ot check this. 

 Apr 15, 2019
 #3
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0

Your answer is right :)

Guest Apr 15, 2019
 #4
avatar+107011 
+1

Good job, Melody  !!!!!!

 

 

cool cool cool

CPhill  Apr 15, 2019
 #6
avatar+107424 
0

Thanks Chris :)

Melody  Apr 16, 2019
 #7
avatar+107424 
0

Hi guest, 

Should I interpret that as 'thanks'   ?

Melody  Apr 16, 2019
 #2
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+1

Melody: These are the numbers I found:
n=1;a= 7*sqrt(-n^2 + 22*n -21); b=n+39;printa,  b,n;n++;if(n<30, goto1,0) 

 

n = 1 , 2, 3, 17, 18, 19, 20, 21.

 Apr 15, 2019
edited by Guest  Apr 15, 2019
 #5
avatar+107424 
0

Your answers are the same as mine.

I think you are just having fun using your C++ programing    wink

Melody  Apr 16, 2019

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