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.
