\[2x ^{1/2}-x ^{1/4}=3\] help

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\[2x ^{1/2}-x ^{1/4}=3\] help

Mathematics
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try to let u=x^1/4 then u^2=x^1/2 so you have 2u^2-u=3 solve for u and then solve for x
2u^2-u-3=0 2u^2-3u+2u-3=0 u(2u-3)+(2u-3)=0 (2u-3)(u+1)=0 u=3/2 u=-1 so x^1/4=3/2 and x^1/4=-1
This will require logarithms. If you rewrite this as 1/2 log 2x - 1/4 log x = 3, this will help. The final answer for x is x=81/16

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x=(3/2)^4=81/16 x=(-1)^4=1 we check these because we raised both sides to an even power
we don't need logarithms
so if we have x=1, does the equation hold? 2(1)-1=3 so 1 is not a solution so if we have x=81/16, does the equation hold? 2(81/16)^(1/2)-(81/16)^(1/4)=2(9/4)-(3/2)=9/2-3/2=6/2=3 so x=81/16 is our only solution
Sorry. I am taking Linear Algebra right now and it has been about 15 years since I took algebra, so I think in terms of logs when I see that. Glad you were able to figure it out. :)

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