anonymous
  • anonymous
How do I solve this? x^2/3 + 4x^1/3 + 3 = 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Factorize it as : (x^(1/3)+1) (x^(1/3)+3) = 0
anonymous
  • anonymous
sorry, wrong didn't see the +3
anonymous
  • anonymous
x never equals zero though... i don't think

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
There are no solutions,I suppose
anonymous
  • anonymous
are you sure you wrote the problem correctly?
anonymous
  • anonymous
Thanks, I'm still confused...I will retype the question...
Xishem
  • Xishem
How do you figure that x never equals 0?
anonymous
  • anonymous
graphing calculator
anonymous
  • anonymous
@ xishem try substituting x = 0
anonymous
  • 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
  • Xishem
\[f(-1)=0\]
anonymous
  • anonymous
no, it equals 2
anonymous
  • anonymous
@studioso33 is it x^(2/3)
anonymous
  • anonymous
IT has no solutions; plot it and see
anonymous
  • anonymous
it has no solution as written in the real numbers, there are non-real answers however
Xishem
  • Xishem
\[f(-1)=x^{2/3}+4x^{1/3}+3=(-1)^{2/3}+4(-1)^{1/3}+3=1-4+3=0\]
anonymous
  • anonymous
SalvatoreTRM, I'm not sure? I've never typed out equations before!
anonymous
  • anonymous
does it look like Xishem wrote it?
anonymous
  • anonymous
x= -27 or -1
Xishem
  • Xishem
I plotted, and there are two solutions.
anonymous
  • anonymous
Sorry, I'm new here...you guys are far more advanced at math than I! Yes, looks like Xishem wrote it...
anonymous
  • anonymous
ok then yes, Ireneoseaspe is right and Xishem as well. x=-27 and -1
anonymous
  • anonymous
Xishem, how to I plot and find solutions myself? Are you all using graphing calculators?
Xishem
  • 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
  • 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
  • anonymous
Ok, i take it back, no solutions. Viva wolfram
anonymous
  • anonymous
Thank you all...it will take time for me to get used to this online tutoring...I will be in here ALOT!
Xishem
  • 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
  • anonymous
http://www.wolframalpha.com/input/?i=cos%5E-1++%28-1%29
anonymous
  • anonymous
who is wolfram?
Xishem
  • 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
  • anonymous
@studio website
anonymous
  • anonymous
If you put x = -1 you get 3 not 0
anonymous
  • 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
  • Xishem
Aron: Show your work. I get 0 every time.
anonymous
  • anonymous
@Irene if you substitute the values you wont get 0.
anonymous
  • anonymous
Xishem, do you know robby mayasich?
anonymous
  • anonymous
nvm, your profile pic reminded me of someone
Xishem
  • Xishem
http://www.wolframalpha.com/input/?i=%28-1%29%5E%281%2F3%29 Wolfram also thinks that is true. Weird.
Xishem
  • Xishem
And that error would explain why WolframAlpha doesn't give the correct answers.
anonymous
  • anonymous
(x^(1/3)+1) (x^(1/3)+3) = 0
anonymous
  • 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.