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Please explain the works of how to solve inequality in n: `C_(2n^2)^(3n+2) >=8`
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Notice that using the factorial formula to express `C_(2n^2)^(3n+2)` is impossible for you to solve the inequality, hence, you should use the following condition such that:
`2n^2 >= 3n+2`
You need to move all terms to one side such that:
`2n^2 - 3n - 2 >= 0`
You need to use quadratic formula to evaluate the zeroes of the expression `2n^2 - 3n - 2 = 0` such that:
`n_(1,2) = (3+-sqrt(9+16))/4 => n_(1,2) = (3+-sqrt25)/4`
`n_(1,2) = (3+-5)/4 => n_1 = 2 ; n_2 = -1/2`
Notice that the expression is positive if `n in [-1/2 ; 2].`
Since you need to keep only the positive integer solutions, hence, evaluating the solutions to the given inequality yields `n=0; n=1; n=2.`
Posted by sciencesolve on September 3, 2012 at 4:13 PM (Answer #1)
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