Question

resolva a equação fatorial (n-2)!/(n)! = 12 :

Answer 1

$\frac{(n-2)!}{(n)!} = 12\\ \frac{(n-2).(n-1).(n)!}{(n)!}=12\\ (n-2).(n-1)=12 \\n^2-n-2n+2=12\\n^2-3n+2-12=0\\n^2-3n - 10 = 0$

a = 1; b = -3; c = -10
equação de segundo grau (considerando D como delta)

$D = b^2 - 4.a.c\\D = (-3)^2 - 4.1.(-10)\\D = 9 + 40\\ D = 49$

$n = \frac{-b +- \sqrt{D} }{2.a}\\\\n' = \frac{-b + \sqrt{D} }{2.a}\\n1 = \frac{-(-3) + \sqrt{49} }{2.1}\\n' = \frac{3 + 7}{2}\\n' = \frac{10}{2}\\n'=5\\\\ n"= \frac{-b - \sqrt{D} }{2.a}\\n" = \frac{-(-3) - \sqrt{49}}{2.1}\\n"= \frac{3 - 7}{2}\\n" = \frac{-4}{2}\\n" = -2$