How do I solve this? x^2/3 + 4x^1/3 + 3 = 0

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How do I solve this? x^2/3 + 4x^1/3 + 3 = 0

Mathematics
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Factorize it as : (x^(1/3)+1) (x^(1/3)+3) = 0
sorry, wrong didn't see the +3
x never equals zero though... i don't think

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Other answers:

There are no solutions,I suppose
are you sure you wrote the problem correctly?
Thanks, I'm still confused...I will retype the question...
How do you figure that x never equals 0?
graphing calculator
@ xishem try substituting x = 0
Solve the equation. x^2/3 + 4x^1/3 + 3 = 0 I think when solving for x, the equation is set equal to zero...
\[f(-1)=0\]
no, it equals 2
@studioso33 is it x^(2/3)
IT has no solutions; plot it and see
it has no solution as written in the real numbers, there are non-real answers however
\[f(-1)=x^{2/3}+4x^{1/3}+3=(-1)^{2/3}+4(-1)^{1/3}+3=1-4+3=0\]
SalvatoreTRM, I'm not sure? I've never typed out equations before!
does it look like Xishem wrote it?
x= -27 or -1
I plotted, and there are two solutions.
Sorry, I'm new here...you guys are far more advanced at math than I! Yes, looks like Xishem wrote it...
ok then yes, Ireneoseaspe is right and Xishem as well. x=-27 and -1
Xishem, how to I plot and find solutions myself? Are you all using graphing calculators?
Why is Wolfram plotting it like this? http://www.wolframalpha.com/input/?i=x%5E%282%2F3%29%2B4x%5E%281%2F3%29%2B3
@studio: I didn't find the original solutions graphically, but rather by factoring and setting each of the factors equal to zero separately.
Ok, i take it back, no solutions. Viva wolfram
Thank you all...it will take time for me to get used to this online tutoring...I will be in here ALOT!
Wolfram is wrong. Take a look at THIS plot, which is just a factorization of the original. http://www.wolframalpha.com/input/?i=%281%2Bx%5E%281%2F3%29%29+%283%2Bx%5E%281%2F3%29%29&lk=1&a=ClashPrefs_*Math- Both the math and my graphing calculator tell me that there are two solutions.
http://www.wolframalpha.com/input/?i=cos%5E-1++%28-1%29
who is wolfram?
\[(x^{1/3}+1) (x^{1/3}+3) = 0\]\[x^{1/3}+1=0 \rightarrow x^{1/3}=-1 \rightarrow x=(-1)^3 \rightarrow x=-1\]You can't deny that there are solutions.
@studio website
If you put x = -1 you get 3 not 0
x^2/3 + 4x^1/3 + 3 = 0 u=x^1/3 u^2 + 4u + 3 = 0 (u + 3)(u + 1) = 0 u= -3,u= - 1 u=x^1/3 -3 = x^1/3 (-3)^3 = (x^1/3)^3 -27 = x u=x^1/3 -1 = x^1/3 (-1)^3 = (x^1/3)^3 -1 = x
Aron: Show your work. I get 0 every time.
@Irene if you substitute the values you wont get 0.
Xishem, do you know robby mayasich?
nvm, your profile pic reminded me of someone
http://www.wolframalpha.com/input/?i=%28-1%29%5E%281%2F3%29 Wolfram also thinks that is true. Weird.
And that error would explain why WolframAlpha doesn't give the correct answers.
(x^(1/3)+1) (x^(1/3)+3) = 0
x^2/3 + 4x^1/3 + 3 = 0 (-27)^2/3 + 4(-27)^1/3 + 3 = 0 \[\sqrt[3]{-27^{2}} + 4\sqrt[3]{-27} + 3 = 0\] \[\sqrt[3]{729} + 4(-3) + 3 = 0\] \[9 -12 + 3 = 0\] \[-3 + 3 =0\] 0=0

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