Hoe los je x ^ (2/3) - 3x ^ (1/3) - 4 = 0 op?

Antwoord:

set # Z = x ^ (1/3) # Wanneer je de # Z # wortels, vind # X = z ^ 3 #

Wortels zijn #729/8# en #-1/8#

Uitleg:

set # X ^ (1/3) = z #

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

Dus de vergelijking wordt:

# 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 #

Op te lossen #X#:

# X ^ (1/3) = z #

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

# X = z ^ 3 #

# X_1 = (9/2) ^ 3 #

# X_1 = 729/8 #

# X_2 = (- 02/01) ^ 3 #

# X_2 = -1/8 #

Antwoord:

x = 64 of x = -1

Uitleg:

Let daar op # (x ^ (1/3)) ^ 2 = x ^ (2/3) #

factoren ontbinden # x ^ (2/3) - 3x ^ (1/3) - 4 = 0 # geeft;

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

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

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

'cubing' beide zijden van het paar vergelijkingen:

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

#rArr x = 64 of x = - 1 #