SOLUTION: I keep getting stomped on how to solve equations like 3x^1/3 + 2x^2/3 = 5 i get it into quadratic formula form but I don't end up getting the right answer.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I keep getting stomped on how to solve equations like 3x^1/3 + 2x^2/3 = 5 i get it into quadratic formula form but I don't end up getting the right answer.      Log On


   

Question 223942: I keep getting stomped on how to solve equations like 3x^1/3 + 2x^2/3 = 5
i get it into quadratic formula form but I don't end up getting the right answer.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Since you do not show your quadratic formula form it is difficult for me to know where you might have gone wrong. But the fact that you are mentioning "quadratic" tells me that you may have understood what may be the hardest part of this problem: that %28x%5E%281%2F3%29%29%5E2+=+x%5E%282%2F3%29. So in this sense the equation is quadratic in terms of x%5E%281%2F3%29. Once you recognize this the rest is relatively easy.

First make one side of this equation equal to zero:
2x%5E%282%2F3%29+%2B+3x%5E%281%2F3%29+-5+=+0
Now we can either factor this or use the quadratic formula (with x%5E%281%2F3%29 = ..., not x = ...):
%28x%5E%281%2F3%29+-+1%29%282x%5E%281%2F3%29+%2B+5%29+=+0
This product can only be zero if one fo the factors is zero so...
x%5E%281%2F3%29+-+1+=+0 or 2x%5E%281%2F3%29+%2B+5+=+0
Solving these:
x%5E%281%2F3%29+=+1 or 2x%5E%281%2F3%29+=+-5
x%5E%281%2F3%29+=+1 or x%5E%281%2F3%29+=+%28-5%29%2F2
Cube each side:
x+=+1 or x+=+%28-125%29%2F8