SOLUTION: Solve for x. 2x^(2/3) - 5x^(1/3) - 3 = 0 Using the Diamond Method, I got the following: 2x^(2/3) - 6x^(1/3) + x^(1/3) - 3 = 0 I then applied factoring by group.

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Solve for x. 2x^(2/3) - 5x^(1/3) - 3 = 0 Using the Diamond Method, I got the following: 2x^(2/3) - 6x^(1/3) + x^(1/3) - 3 = 0 I then applied factoring by group.      Log On


   

Question 1199265: Solve for x.

2x^(2/3) - 5x^(1/3) - 3 = 0

Using the Diamond Method, I got the following:

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

I then applied factoring by group.

2x^(1/3)[x^(1/3) - 3]

Setting each factor to 0 and solving for x, I got the following

values: 0, 27

According to the textbook, x = 27 is one of the solutions.

The second value of x should be -1/8.

I did everything I could think of to find the second value of x but to no avail.

How do I find x = -1/8?


Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve for x     2x^(2/3) - 5x^(1/3) - 3 = 0
~~~~~~~~~~~~~~~~~ 

Prepare your equation to factoring using grouping-regrouping  (in the same way as you did in your post)

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


Now factor left side step by step

    2x^(1/3)*(x^(1/3)-3) + (x^(1/3) - 3) = 


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


From this factored form you have EITHER first parentheses expression is zero, 
OR the second parentheses expression is zero.


(a)  Case  x^(1/3)-3 = 0.

     It gives  x^(1/3) = 3,  which implies  x = 3^3 = 27.



(b)  Case  2x^(1/3)+1 = 0.

     It gives  2x^(1/3) + 1  0,  which implies  x^(1/3) = -1%2F2,  x = %28-1%2F2%29%5E3 = -1%2F8.

Solved.

The answers/(the solutions)   are   x= 27   and/or   x= -1%2F8,   same as in your textbook.

Victory (!)


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In your post, your very first step was correct, but your factoring after that was wrong.