Question

Persamaan x3 - 2x² + 6x - 3 = 0 memiliki akar - akar x1,x2, dan x3. Tentukan nilai dari 1/x1 + 1/x2 + 1/x3?

Answer 1

Suku banyak

x³ - 2x² + 6x - 3 = 0
a = 1 , b = - 2 , c = 6 , d = - 3

x1+x2+x3 = -b/a = 2
x1x2 + x2 x3 + x1 x3 = c/a = 6
x1 x2 x3 = -d/a = 3

1/x1 + 1/x2 + 1/x3 = {(x2+x3) + (x1 x3) + (x1 x2)}/ ( x1 x2 x3)
1/x1 +1/x2 +1/x3 = (6/3) = 2