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

1 Answer

Answer:

#x=8/27# and #x=-1#

Explanation:

From the given equation
#3x^(2/3)+x^(1/3)-2=0#

The solution goes like this

Let #w=x^(1/3)#

#3x^(2/3)+x^(1/3)-2=0#

we can write this equation this way

#3(x^(1/3))^2+(x^(1/3))^1-2=0#

also

#3(w)^2+1*(w)^1-2=0#

also let #a=3# and #b=1# and #c=-2#

#w=(-b+-sqrt(b^2-4ac))/(2a)#

#w=(-1+-sqrt(1^2-4*3*(-2)))/(2*3)#

#w=(-1+-sqrt(1+24))/6#

#w=(-1+-sqrt25)/6#

#w=(-1+-5)/6#

we have two values for w:

#w_1=(-1+5)/6=4/6=2/3# and #w_2=(-1-5)/6=-6/6=-1#

But #w=x^(1/3)#

and #x=w^3#

Using #w=2/3#

#x=w^3=(2/3)^3=8/27#
#x=8/27#

Using #w=-1#

#x=w^3=(-1)^3=-1#

#x=-1#

check at #x=8/27#
#3x^(2/3)+x^(1/3)-2=0#
#3(8/27)^(2/3)+(8/27)^(1/3)-2=0#
#3(4/9)+2/3-2=0#
#4/3+2/3-2=0#
#6/3-2=0#
#2-2=0#
#0=0#
#x=8/27# is a root

check at #x=-1#
#3x^(2/3)+x^(1/3)-2=0#
#3(-1)^(2/3)+(-1)^(1/3)-2=0#
#3(1)-1-2=0#
#3-3=0#
#0=0#

#x=-1# is a root

God bless....I hope the explanation is useful.