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$?
$\endgroup$ 03 Answers
$\begingroup$This is still a hint
$$x^{2/3}=4$$ $$x^{2}=4^3=64$$
$$x=\pm8$$
I leave the rest to you!
$\endgroup$ 5 $\begingroup$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 $x^{(2/3)}$
$\endgroup$ 4 $\begingroup$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 $\pm 8$ 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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