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

- anonymous

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

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- anonymous

Factorize it as : (x^(1/3)+1) (x^(1/3)+3) = 0

- anonymous

sorry, wrong didn't see the +3

- anonymous

x never equals zero though... i don't think

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## More answers

- anonymous

There are no solutions,I suppose

- anonymous

are you sure you wrote the problem correctly?

- anonymous

Thanks, I'm still confused...I will retype the question...

- Xishem

How do you figure that x never equals 0?

- anonymous

graphing calculator

- anonymous

@ xishem try substituting x = 0

- anonymous

Solve the equation.
x^2/3 + 4x^1/3 + 3 = 0
I think when solving for x, the equation is set equal to zero...

- Xishem

\[f(-1)=0\]

- anonymous

no, it equals 2

- anonymous

@studioso33 is it x^(2/3)

- anonymous

IT has no solutions; plot it and see

- anonymous

it has no solution as written in the real numbers, there are non-real answers however

- Xishem

\[f(-1)=x^{2/3}+4x^{1/3}+3=(-1)^{2/3}+4(-1)^{1/3}+3=1-4+3=0\]

- anonymous

SalvatoreTRM, I'm not sure? I've never typed out equations before!

- anonymous

does it look like Xishem wrote it?

- anonymous

x= -27 or -1

- Xishem

I plotted, and there are two solutions.

- anonymous

Sorry, I'm new here...you guys are far more advanced at math than I! Yes, looks like Xishem wrote it...

- anonymous

ok then yes, Ireneoseaspe is right and Xishem as well. x=-27 and -1

- anonymous

Xishem, how to I plot and find solutions myself? Are you all using graphing calculators?

- Xishem

Why is Wolfram plotting it like this?
http://www.wolframalpha.com/input/?i=x%5E%282%2F3%29%2B4x%5E%281%2F3%29%2B3

- Xishem

@studio: I didn't find the original solutions graphically, but rather by factoring and setting each of the factors equal to zero separately.

- anonymous

Ok, i take it back, no solutions. Viva wolfram

- anonymous

Thank you all...it will take time for me to get used to this online tutoring...I will be in here ALOT!

- Xishem

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.

- anonymous

http://www.wolframalpha.com/input/?i=cos%5E-1++%28-1%29

- anonymous

who is wolfram?

- Xishem

\[(x^{1/3}+1) (x^{1/3}+3) = 0\]\[x^{1/3}+1=0 \rightarrow x^{1/3}=-1 \rightarrow x=(-1)^3 \rightarrow x=-1\]You can't deny that there are solutions.

- anonymous

@studio
website

- anonymous

If you put x = -1 you get 3 not 0

- anonymous

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

- Xishem

Aron: Show your work. I get 0 every time.

- anonymous

@Irene if you substitute the values you wont get 0.

- anonymous

Xishem, do you know robby mayasich?

- anonymous

nvm, your profile pic reminded me of someone

- Xishem

http://www.wolframalpha.com/input/?i=%28-1%29%5E%281%2F3%29
Wolfram also thinks that is true. Weird.

- Xishem

And that error would explain why WolframAlpha doesn't give the correct answers.

- anonymous

(x^(1/3)+1) (x^(1/3)+3) = 0

- anonymous

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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