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Find the number of integers n that satisfy

\(7\sqrt{-n^2+22n-21}≤n+39\)

 Apr 7, 2021
edited by imbadatmath  Apr 7, 2021
 #1
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The expression \(-n^2+22n-21\) will only be positive if \(1\leq x\leq21\). (If it's negative then the left-hand side is undefined), which means that both sides of the expression must be positive, so we can square both sides without worrying about changing signs.

\(49(-n^2+22n-21)\leq n^2+78n+1521\\-49n^2+1078n-1029\leq n^2+78n+1521\\ 0\leq50n^2-1000n+2550\\0\leq n^2-20n+51\\0\leq (x+3)(x+17)\)

The zeroes of the quadratic of the left-hand side are -3 and -17, which means that the answer is \(\boxed{15}\)

 Apr 7, 2021
 #2
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Revwriting the previous answer:

 

\(-n^2 + 22n - 21 = -(n^2 - 22n + 21) = - (n-21)(n-1)\)

 

For the above to be non-negative, as stated in the previous solution:

 

\(1 \le n \le 21\)

 

Therefore we are starting with a possibility of 21 integers.

 

From the previous answer, the following was derived:

 

\(0 \le n^2 - 20n + 51\)

 

Factoring this yields:

 

\(0 \le (n-3)(n-17)\)

 

For this to be true:

 

\(3 \le n \le 17\)

 

The number of integers satisfying this is 15.

 Apr 7, 2021
edited by Guest  Apr 7, 2021

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