OK I know this sounds pretty stupid, but I am stuck on solving $x^{{2}/{3}}=4$. I rewrote it to $\sqrt[3]{x^2}=4$, but I don't know what to do next. Would the radical go away if I took the $\sqrt[3]{x^2}=4$ by the $3$rd power?

Then it would become $ x^6=64$?


This is still a hint

$$x^{2/3}=4$$ $$x^{2}=4^3=64$$


I leave the rest to you!

  • $\begingroup$ omg thank you so much! I totally forgot! $\endgroup$ – Elsa Dec 7 '14 at 6:49
  • $\begingroup$ so if the domain of this is [-1,27] then that means that -8 is not an answer, right? $\endgroup$ – Elsa Dec 7 '14 at 6:53
  • $\begingroup$ @Amzoti I think -8 is part of the original solution???? since it's x^2 $\endgroup$ – Elsa Dec 7 '14 at 6:54
  • $\begingroup$ @Amzoti according to the calculator it's 4. and 4-4=0. $\endgroup$ – Elsa Dec 7 '14 at 6:57
  • 3
    $\begingroup$ @Kaaagome: raising a negative number to a non-integral power is problematic in the reals. Often it is not defined-you need to check your definition. Rational powers with an odd denominator can work, defied by the usual rules of exponents. $\endgroup$ – Ross Millikan Dec 7 '14 at 7:07

You multiply both sides by $3$, getting $x^2=12$. Can you take it from here? Note that x^2/3 usually means $(x^2)/3$, not $x^{(2/3)}$

  • 2
    $\begingroup$ Down-vote should be reversed but till then +1 for compensation! $\endgroup$ – Aditya Hase Dec 7 '14 at 6:52
  • $\begingroup$ His question was edited and this isn't really an answer to the question $\endgroup$ – Joao Dec 7 '14 at 6:56
  • $\begingroup$ Before the edit, this was the only correct answer. +1 for you. OP (and other readers) should take this as a chastisement for poor notation. Order of operations is quite important and parentheses simply must be used if it is to be altered. $\endgroup$ – MPW Dec 7 '14 at 7:01
  • $\begingroup$ @Joao: I answered before the edit. I think parentheses are important and this is a good example. Yes, it is likely that x^2/3 is intended as $x^{2/3}$, but the order of operations says otherwise. $\endgroup$ – Ross Millikan Dec 7 '14 at 7:04

note that if two positive quantities are equal, then you can raise them to the same power then the result will be equal.

So you cube both sides, to get x^2=64. There we get +-8 as solutions.

Clearly x=8 works (8^{2/3}=4) On the other hand (-8)^{2/3}=4, so both 8 and -8 are solutions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.