How do you solve #x^(2/3) - 3x^(1/3) - 4 = 0#?

2 Answers
Costas C.
Jan 14, 2016

Set #z=x^(1/3)# When you find the #z# roots, find #x=z^3#

Roots are #729/8# and #-1/8#

Explanation:

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

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

So the equation becomes:

#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#

To solve for #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#

Jim G.
Jan 14, 2016

x = 64 or x = -1

Explanation:

note that # (x^(1/3))^2 = x^(2/3) #

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

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

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

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

'cubing' both sides of the pair of equations :

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

#rArr x = 64 or x = - 1 #

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