Kako rešiti x ^ (2/3) - 3x ^ (1/3) - 4 = 0?

Odgovor:

Set # z = x ^ (1/3) # Ko najdete # z # korenine, poiščite # x = z ^ 3 #

Korenine so #729/8# in #-1/8#

Pojasnilo:

Set # x ^ (1/3) = z #

# x ^ (2/3) = x ^ (1/3 * 2) = (x ^ (1/3)) ^ 2 = z ^ 2 #

Enačba postane:

# z ^ 2-3z-4 = 0 #

# Δ = b ^ 2-4ac #

#Δ=(-3)^2-4*1*(-4)#

#Δ=25#

#z_ (1,2) = (- b + -sqrt (Δ)) / (2a) #

#z_ (1,2) = (- (- 4) + - sqrt (25)) / (2 * 1) #

#z_ (1,2) = (4 + -5) / 2 #

# z_1 = 9/2 #

# z_2 = -1 / 2 #

Rešiti za # x #:

# x ^ (1/3) = z #

# (x ^ (1/3)) ^ 3 = z ^ 3 #

# x = z ^ 3 #

# x_1 = (9/2) ^ 3 #

# x_1 = 729/8 #

# x_2 = (- 1/2) ^ 3 #

# x_2 = -1 / 8 #

Odgovor:

x = 64 ali x = -1

Pojasnilo:

Upoštevajte, da # (x ^ (1/3)) ^ 2 = x ^ (2/3) #

Faktorizacija # x ^ (2/3) - 3x ^ (1/3) - 4 = 0 # daje;

# (x ^ (1/3) - 4) (x ^ (http: // 3) + 1) = 0 #

#rArr (x ^ (1/3) - 4) = 0 ali (x ^ (1/3) + 1) = 0 #

#rArr x ^ (1/3) = 4 ali x ^ (1/3) = - 1 #

"kubiranje" obeh strani enačb:

# (x ^ (1/3)) ^ 3 = 4 ^ 3 in (x ^ (1/3)) ^ 3 = (- 1) ^ 3 #

#rArr x = 64 ali x = - 1 #