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I'm having difficulty solving the following equation for $x$:

$$\sqrt[m]{(1+x)^2}-\sqrt[m]{(1-x)^2}=\sqrt[m]{1-x^2}$$

I have tried a few substitutions but that didn't seem to get me anywhere. Can someone please help me with that? Thanks in advance.

daniels
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    Try dividing by $\sqrt[m]{(1-x)^2}$ and putting $y=\frac{1+x}{1-x}$, or even $y=\sqrt[m]{\frac{1+x}{1-y}}$. – OR. Jul 01 '17 at 02:22

1 Answers1

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Let $u = (1+x)^{1/n},$ and $v = (1-x)^{1/n}.$ Then, your equation is:

$$u^2 - v^2 = u v.$$ Now, letting $w = u/v,$ we get

$$w - 1/w = 1,$$ which is a quadratic equation in $w$ (in fact, the solution is $w = \varphi,$ the golden ratio.

Now, $$\frac{1+x}{1-x}=\varphi^n,$$ so you have a linear equation for $x.$

Igor Rivin
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  • Thanks for your answer. I've worked that linear equation out and found $x$ to be equal to $(φ^n-1)/(φ^n+1)$. However, my book says there should be two answers if $n$ is odd and one answer in case it is even. Can you please help me understand that? Thanks. – daniels Jul 01 '17 at 04:10
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    @daniels A quadratic equation has two solutions - find the other, and see what happens... – Igor Rivin Jul 01 '17 at 15:20
  • So, I have indeed found two solutions, namely $x_1=\frac{(\frac{1+\sqrt{5}}{2})^n-1}{(\frac{1+\sqrt{5}}{2})^n+1}$ and $x_2=\frac{(\frac{1-\sqrt{5}}{2})^n-1}{(\frac{1-\sqrt{5}}{2})^n+1}$. What I still don't understand (and this may be a really silly question, but I really can't grasp it) is why only one (and which one) of them applies when $n$ is even. Thanks for your patience. – daniels Jul 01 '17 at 15:29
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    @daniels You can't take an even root of a negative number, but you can take an odd root. Does that help? – Igor Rivin Jul 01 '17 at 15:35
  • Hum, I know when you raise both sides of an equation to an even power this may produce extraneous solutions. I was wondering whether it had anything to do with that. As to your response, $(1+x)^2$ and $(1-x)^2$ are always positive no matter the value of $x$, and the fact that $x$ is being raised to the second power, in the right term of the equation, also assures it will always be positive, right? So I still can't see how it makes a difference, as (or so it appears to me) we'd be taking a root of a positive number in either case. – daniels Jul 01 '17 at 16:01
  • After giving it some thought, I think I've figured out the answer. For $x_2$, the left hand side of the original equation is always negative, no matter the value of $n$. However, when $n$ is even, $|x_2|<1$ and hence $1-(x_2)^2>0$, which causes the right hand side to be positive. What intrigues me, though, is whether there was an easier way to somehow anticipate this result. I mean, if it weren't for the book I would have never suspected one of the solutions doesn't work for some choices of $n$. – daniels Jul 01 '17 at 22:37
  • Is there some tip you can give me as to when to expect such strange results and, if this is possible, how to identify them without having to test every solution that is found? Thanks anyway, you have been a great help. – daniels Jul 01 '17 at 22:42
  • @daniels You are welcome, but in general you just have to be careful - just do lots of examples, and you will be used to checking things. – Igor Rivin Jul 01 '17 at 23:29