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

1. anonymous

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

2. anonymous

sorry, wrong didn't see the +3

3. anonymous

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

4. anonymous

There are no solutions,I suppose

5. anonymous

are you sure you wrote the problem correctly?

6. anonymous

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

7. Xishem

How do you figure that x never equals 0?

8. anonymous

graphing calculator

9. anonymous

@ xishem try substituting x = 0

10. 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...

11. Xishem

$f(-1)=0$

12. anonymous

no, it equals 2

13. anonymous

@studioso33 is it x^(2/3)

14. anonymous

IT has no solutions; plot it and see

15. anonymous

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

16. Xishem

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

17. anonymous

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

18. anonymous

does it look like Xishem wrote it?

19. anonymous

x= -27 or -1

20. Xishem

I plotted, and there are two solutions.

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. 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

25. Xishem

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

26. anonymous

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

27. anonymous

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

28. 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.

29. anonymous
30. anonymous

who is wolfram?

31. 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.

32. anonymous

@studio website

33. anonymous

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

34. 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

35. Xishem

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

36. anonymous

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

37. anonymous

Xishem, do you know robby mayasich?

38. anonymous

nvm, your profile pic reminded me of someone

39. Xishem

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

40. Xishem

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

41. anonymous

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

42. 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