SOLUTION: Hello! My problem is finding all real solutions with the following equation {{{x^(4/3)-5x^(2/3)+6}}} I put 6 on the other side so it turns into {{{x^(4/3)-5x^(2/3)=-6}}} I can'

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Hello! My problem is finding all real solutions with the following equation {{{x^(4/3)-5x^(2/3)+6}}} I put 6 on the other side so it turns into {{{x^(4/3)-5x^(2/3)=-6}}} I can'      Log On


   

Question 352103: Hello!
My problem is finding all real solutions with the following equation x%5E%284%2F3%29-5x%5E%282%2F3%29%2B6
I put 6 on the other side so it turns into
x%5E%284%2F3%29-5x%5E%282%2F3%29=-6
I can't combine the two terms so I ended up subtracting -5x^(2/3) from both sides to get:
x%5E%284%2F3%29=5x%5E%282%2F3%29-6
Then I will take the power 3/4th to both sides so X on the left is by itself.
x=%285x%5E%282%2F3%29%29%5E%283%2F4%29-6%5E%283%2F4%29
Which will simplify to:
125%5E%281%2F4%29%2Ax%5E%281%2F2%29-126%5E%281%2F4%29=X
This is where I get stuck.
I have a feeling I screwed up somewhere.
Can you please help me?
Thank you.
-Kayte.
Found 2 solutions by Fombitz, stanbon:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
No, the key to this problem is to recognize that it's a quadratic equation in disguise.
Use a substitution.
Let u=x%5E%282%2F3%29, then u%5E2=x%5E%284%2F3%29, then
x%5E%284%2F3%29-5x%5E%282%2F3%29%2B6=0 becomes
u%5E2-5u%2B6=0
Then you solve for the solutions to u by factoring, completing the square, or quadratic formula.
Then once you have u, back substitute and find x.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
My problem is finding all real solutions with the following equation x%5E%284%2F3%29-5x%5E%282%2F3%29%2B6
I put 6 on the other side so it turns into
x%5E%284%2F3%29-5x%5E%282%2F3%29=-6
I can't combine the two terms so I ended up subtracting -5x^(2/3) from both sides to get:
x%5E%284%2F3%29=5x%5E%282%2F3%29-6
Then I will take the power 3/4th to both sides so X on the left is by itself.
x=%285x%5E%282%2F3%29%29%5E%283%2F4%29-6%5E%283%2F4%29
Which will simplify to:
125%5E%281%2F4%29%2Ax%5E%281%2F2%29-126%5E%281%2F4%29=X
============================================
x^(4/3)-5x^(2/3)+6}
Your problem is a quadratic with variable x^(2/3).
---------------------------
It can be written as:
[x^(2/3)]^2 - 5[x^(2/3)] +6 = 0
---
It factors as:
[x^(2/3)-3][x^(2/3)-2] = 0
Then:
x^(2/3) = 3 or x^(2/3) = 2
x = 3^(3/2) or x = 1^(3/2)
---
x = 3sqrt(3) or x = 1
===========================
Cheers,
Stan H.