Paano mo malutas ang x ^ (2/3) - 3x ^ (1/3) - 4 = 0?

Sagot:

Itakda # z = x ^ (1/3) # Kapag nakita mo ang # z # mga ugat, hanapin # x = z ^ 3 #

Ang mga ugat ay #729/8# at #-1/8#

Paliwanag:

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

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

Kaya ang equation ay nagiging:

# 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 #

Upang malutas ang # 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 #

Sagot:

x = 64 o x = -1

Paliwanag:

tandaan na # (x ^ (1/3)) ^ 2 = x ^ (2/3) #

Factorising # x ^ (2/3) - 3x ^ (1/3) - 4 = 0 # nagbibigay;

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

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

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

'cubing' sa magkabilang panig ng pares ng equation:

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

#rArr x = 64 o x = - 1 #