Hi, I am stuck on this maths problem. Solve for n: (n+2)!/(n)! =110. I get as far as (n+2)(n+1) = 110. When I multiply it out I get n^2,3n,2 = 110. There is only one example similar to this in my book and it used a quadratic equation to solve it. I feel like that is the next step. I am just unsure of what to do with the 2 I have left over from multiplying out the binomial. I'm really sorry if this doesn't make sense. It's been several years since I had to study mathematics and I am really rusty. Any help is greatly appreciated.
-
So you have n^2 + 3n + 2 = 110 ----> n^2 + 3n - 108 = 0 ----> (n + 12)(n - 9) = 0.
So either n = -12 or n = 9, but since we require n > 0 we have n = 9 as the solution.
Check: (9 + 2)! / 9! = 11! / 9! = 11*10 = 110.
So either n = -12 or n = 9, but since we require n > 0 we have n = 9 as the solution.
Check: (9 + 2)! / 9! = 11! / 9! = 11*10 = 110.
-
(n+2)(n+1) = 110
n^2 + 3n +1 = 110...set equal to zero.....subtract 110
n^2 +3n - 109 = 0
quadratic equation
x = [-b +/- sqrt(b^2 - 4ac)} / 2a
x = -3/2 +/- sqrt( 3^2 + 436)/2
calculator
n^2 + 3n +1 = 110...set equal to zero.....subtract 110
n^2 +3n - 109 = 0
quadratic equation
x = [-b +/- sqrt(b^2 - 4ac)} / 2a
x = -3/2 +/- sqrt( 3^2 + 436)/2
calculator
-
n² + 3n + 2 = 110
n² + 3n - 108 = 0
(n + 12)(n - 9) = 0
n = - 12 , n = 9
n² + 3n - 108 = 0
(n + 12)(n - 9) = 0
n = - 12 , n = 9