Question

O valor de n na equação n!/(n-2)!=42 é:
A)1
B)3
C)5
D)7
E)11

Answer 1

Explicação passo a passo:

$\sf \dfrac{n!}{(n-2)!} = 42$

$\sf \dfrac{n*(n-1)*(n-2)!}{(n-2)!} = 42$

$\sf \dfrac{n*(n-1)*\cancel{(n-2)!}}{\cancel{(n-2)!}} = 42$

$\sf n*(n - 1) = 42$

$\sf n^2 - n = 42$

$\sf n^2 - n - 42 = 0$

Coeficientes: a = 1, b = - 1, c = 0

$\sf n = \dfrac{- b \pm \sqrt{b^2 - 4ac}}{2a}$

$\sf n = \dfrac{- (-1) \pm \sqrt{(-1)^2 - 4"(1)*(-42)}}{2*(1)}$

$\sf n = \dfrac{1 \pm \sqrt{1 + 168}}{2}$

$\sf n = \dfrac{1 \pm \sqrt{169}}{2}$

$\sf n = \dfrac{1 \pm 13}{2}$

•$\sf ~~n' = \dfrac{1 + 13}{2} = \dfrac{14}{2} =\red{ 7}$

•$\sf ~~n'' = \dfrac{1 - 13}{2} = - \dfrac{12}{2} = \red{- 6}$

Encontramos duas raizés

n' = 7

n'' = - 6

Considerando apenas a raiz positiva, n = 7

=====> Letra (D)