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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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drifterkay | Student, Undergraduate | (Level 2) eNoter

Posted September 3, 2012 at 4:04 PM via web

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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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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 3, 2012 at 4:13 PM (Answer #1)

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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.`

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