It might seem very stupid question.
If $x^2=9$ then to solve for $x$ we take both principal $n$-th root of $9$, i.e. $3$ and the negative $n$-th root of $9$, i.e. $-3$. This is right until I found about rational exponents.
If I try to solve the same equation using rational exponents, then $x^2=9$ will be $(x^2)^{1/2} =9^{1/2}$ which leads to $x= 3^{2\cdot (1/2)}$. Here solution will only be $x=3$ the principal $2$-nd root only.
I know the $x$ should equal $\pm3$. What am I doing wrong solving equation using rational exponents?
Best Answer
The problem is that $(x^2)^{1/2} = \lvert x\rvert$, not $x$. So $\lvert x \rvert = 3$ and therefore $x = \pm 3$.