Question

Jika x₁ dan x₂ adalah akar-akar persamaan log^2 (x + 2) + log (x + 2)^3 = log 0,01 dengan x₁ > x₂, maka nilai dari x₁ - x₂ = ….

Answer 1

Log² (x + 2) + log (x + 2)³ = log 10^-2
log² (x + 2) + 3 log (x + 2) + 2 = 0
(log (x + 2) + 1) (log (x + 2) + 2) = 0

log (x + 2) + 1 = 0
log (x + 2) = -1
x + 2 = 10^-1 = 1/10
x = 1/10 - 2
x = -19/10

log (x + 2) + 2 = 0
log (x + 2) = -2
x + 2 = 10^-2
x = 1/100 - 2
x = -199/100

x1 > x2
x1 = -19/10 = -1,9
x2 = -199/100 = -1,99

x1 - x2 = -1,9 - (-1,99) = 0,09