121/n^2>8 subtract 8 from both sides
121/n^2 - 8 > 0 simplify
[ 121 - 8n^2] / n^2 > 0 factor the LHS
0 < n < 11/(2 sqrt(2)), which is your answer.
+23254 121 / n2 > 8
Multiply both sides by n2; since n2 > 0, you don't have to change the direction of the inequality.
121 / n2 > 8 ---> 121 > 8n2 ---> 121/8 > n2 ---> n2 < 121/8
This means that: - sqrt(121/8) < n < sqrt(121/8) (with n not equal to zero)
Or: - [11 · sqrt(2)] / 4 < n < [11 · sqrt(2)] / 4
