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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$?

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This is still a hint

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

$$x=\pm8$$

I leave the rest to you!

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $x^{(2/3)}$

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    $\begingroup$ 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
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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.

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