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$$ \begin{align} 2z &= 1-\sqrt{x} \\ 2y &= 1-\sqrt{z} \\ 2x &= 1-\sqrt{y} \end{align} $$ Is there a way to solve it? I have tried using simultaneous equations, but it ends up with 0=0. I have tried squaring both sides and getting rid of one variable, but it just ends up with 0=0 in the end, is it that there is multiple real solutions? I'm confused because the question states to find all real solutions. Is it a function or is there actually solutions?

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  • $\begingroup$ Hi, welcome to Math.SE. Please indicate what you have tried and where you are stuck. (i.e. how did you get to $0=0$?) This will help people better tailor their answer to your background and situation. It will also demonstrate that you are interested in your question and not just looking for someone to do your homework for you - Math.SE is not a homework site. $\endgroup$ – Ian Miller May 1 '16 at 8:29
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Eliminate the square roots by rewriting

$$x=(1-2z)^2,\\y=(1-2x)^2,\\z=(1-2y)^2.$$

Then by substitution you obtain a nice polynomial equation,

$$x=(1-2z)^2=(1-2(1-2y)^2)^2=(1-2(1-2(1-2x)^2)^2)^2$$

that has $8$ real roots in $[0,1]$, among which $\frac14$ and $1$.

enter image description here

All these roots are valid solutions, as they yield positive RHS.


Better, if you set $x=\cos^2(w)$, the equation is

$$\cos(u)=1-2\cos^2(w)=-\cos(2w),\\ \cos(v)=1-2\cos^2(2w)=-\cos(4w),\\ \cos(w)=1-2\cos^2(2w)=-\cos(8w),$$

and

$$w=\pm(\pi-8w)+2k\pi.$$ For completeness, the signs must be discussed.

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