Question

A soma das raízes da equação [(n-1)!]^2 -7 * (n - 1)! + 6 = 0

Answer 1

$(n-1)! = x \implies x^2-7x+6\therefore \Delta=49-4*1*6=49-24=25=5^2\therefore x = \frac{7 \pm 5}{2}\implies x_1=6, e , x_2=1$

$(n-1)!=6\therefore (n-1)!=3*2*1\therefore n-1=3\implies n=4\\(n-1)!=1\therefore n-1=0 , ou, n-1=1\therefore n=1, ou, n=2$

Somando $2+4+1=7$