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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?

Gibbs
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ahmed allam
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1 Answers1

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The problem is that $(x^2)^{1/2} = \lvert x\rvert$, not $x$. So $\lvert x \rvert = 3$ and therefore $x = \pm 3$.

Gibbs
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  • +1 if this is documented at any text please show it to me to accept the answer.i mean any text saying thar x^a^(1/a)=|x| – ahmed allam May 29 '20 at 09:16
  • I am not sure I understand your comment... In any case, the square root of a (non-negative) number $x$ is by definition that non-negative number $y$ such that $y^2 = x$. Since $x$ might be negative, you need to make sure that $(x^2)^{1/2}$ is non-negative, hence the result is $\lvert x\rvert$. – Gibbs May 29 '20 at 09:21
  • in the book i am reading :((x)^a)^(1\a)=x^a*(1/a)=x^1=x not|x| the book is ALGEBRA 2 and TRIGONOMETRY amasco page 304 – ahmed allam May 29 '20 at 09:26
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    I guess saying that $x^{a \cdot 1/a} = x$ the author considers all possible solutions. For example, $((-2)^3)^{1/3}$ has a unique solution $-2$, but $((-2)^4)^{1/4} = \pm 2$. It might as well be that the author does not bother to be too precise on this point. – Gibbs May 29 '20 at 09:33
  • this is a book with your definition:College Algebra and Trigonometry By Richard N. Aufmann, Vernon C. Barker, – ahmed allam May 30 '20 at 09:54