Find the number of integers n that satisfy

\(7\sqrt{-n^2+22n-21}≤n+39\)

imbadatmath Apr 7, 2021

#1**0 **

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 \(\boxed{15}\)

textot Apr 7, 2021

#2**+1 **

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.

Guest Apr 7, 2021

edited by
Guest
Apr 7, 2021