SOLUTION: 1.) Find the solutions of the equation. I already subuted u for x^(2/3) then i got a quadratic equation 3u^2 + 5u - 2=0 and i got u = 1/3 and -2 and i have to sub somethin

Question 789099: 1.) Find the solutions of the equation.
I already subuted u for x^(2/3) then i got a quadratic equation 3u^2 + 5u - 2=0 and i got u = 1/3 and -2 and i have to sub something back but I do not know what to do now after i got those two answers for u?
3x^(4/3) + 5x^(2/3) - 2=0
Can someone please help me on this problem and go step by step with this equtaion for me? THANK YOU!

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
You let

So because the first solution for 'u' is , this means...

or

or

or

or

or

or

or

or

or

-------------------------------------------------------

Now if , then...

or

or

or

or

or

Since these solutions are complex and not real, we can ignore them.

=======================================================

So the two *possible* real numbered solutions are

or

It turns out that when you plug each back into the original equation, only works while makes the original equation false.

Therefore, the only solution is
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