Question 228412: 20x^(2/3) - 6x^(1/3) - 2 = 0
I have tried using my book's suggestion to solve this problem, which is to use a variable, in this case, for (x^1/3) = m. What winds up happening is that I wind up with a factorization problem that doesn't make sense (unfortunately, the text book shows us only the SIMPLEST form of a certain type of problem, leaving you on your own when it comes time to solve something a little more challenging). I am having difficulty grasping the steps and proper execution of this problem and would greatly appreciate your help. Thank you.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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