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'

Question 352103: Hello!
My problem is finding all real solutions with the following equation
I put 6 on the other side so it turns into

I can't combine the two terms so I ended up subtracting -5x^(2/3) from both sides to get:

Then I will take the power 3/4th to both sides so X on the left is by itself.

Which will simplify to:

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) (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 , then , then
becomes

Then you solve for the solutions to by factoring, completing the square, or quadratic formula.
Then once you have , back substitute and find .
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
My problem is finding all real solutions with the following equation
I put 6 on the other side so it turns into

I can't combine the two terms so I ended up subtracting -5x^(2/3) from both sides to get:

Then I will take the power 3/4th to both sides so X on the left is by itself.

Which will simplify to:

============================================
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.
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