Question 228412
Your equation to solve is:


20x^(2/3) - 6x^(1/3) - 2 = 0 


If you let y = x^(1/3), then y^2 = x^(1/3)^2 = x^(2/3)


Your equation would become:


20y^2 - 6y - 2 = 0


This might be able to be factored to something like:


(5y + 1) * (4y - 2)


To test this out, do the multiplications of the factors as shown below:


-2 * 1 = -2
-2 * 5y = -10y
4y * 1 = 4y
4y * 5y = 20y

Add these together and combine like terms and you get:


20y - 10y + 4y - 2 becomes:
20y - 6y - 2


So you have found the factors of:


(5y + 1) * (4y - 2) = 0


This results in:


5y + 1 = 0 which results in y = -.2
4y - 2 = 0 which results in y = .5


Now you had made y = x^(1/3) so this equation becomes:


x^(1/3) = -.2
and:
x^(1/3) = .5


x^(2/3) = x^(1/3)^2 = -.2^ = .04
and:
x^(2/3) = x^(1/3)^2 = .5^ = .25


We hold these values to test against the original equation.


First values are x^(1/3) = -.2 and x^(2/3) = .04


Plug into the original equation of:


20x^(2/3) - 6x^(1/3) - 2 = 0 to get:


20*.04 - 6*(-.2) - 2 = 0
simplify to get:
.8 + 1.2 - 2 = 0
combine like terms to get:
0 = 0 which is true so it looks like our first values for x^(1/3) and x^(2/3) are good.


Our second values are x^(1/3) = .5 and x^(2/3) = .25


Plug into original equation of :


20x^(2/3) - 6x^(1/3) - 2 = 0 to get:


20*.25 - 6*.5 - 2 = 0
simplify to get:
5 - 3 - 2 = 0
combine like terms to get:
0 = 0


Looks like our second value for x^(1/3) and x^(2/3) are also good.


Your answer are that:


x^(1/3) = -.2
and:
x^(1/3) = .5


To solve for x, you would need to cube both sides of this equation to get:


x = (-.2)^3 = -.0008
and:
x = (.5)^3 = .125


You can use your calculator to prove that these values are good.


Your original equation is:


20x^(2/3) - 6x^(1/3) - 2 = 0 


If x = -.008, then this equation becomes:


20*(-.008)^(2/3) - 6*(-.008)^(1/3) - 2 = 0


This  becomes:


20* ((-.008)^2)^(1/3) - 6*(-.008)^(1/3) - 2 = 0


-.008^2 = .000064^(1/3) = .04
-.008^(1/3) = -.2


Equation becomes 20 * .04 - 6 * (-.2) - 2 = 0


This becomes .8 - (-1.2) - 2 = 0


This becomes .8 + 1.2 - 2 = 0


This becomes 2 - 2 = 0 which is true so the first value for x is good.


I'll leave the confirmation of the second value of x to you.


You're right.


It was a bear of a problem to solve.


The key was understanding that y = x^(1/3) and that y^2 equal x^(2/3).


This was one of the properties of exponents.


(x^a)^2 = x^a * x^a = x^(a+a) = x^(2a)


Even knowing this, the problem was difficult.