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Hi jacks,Calculate $x$ in $\{x^2\}+\{x\} = 1$
where $\{x\} = $ fractional part of $x$
The first solution does not satisfy the original equation. I'm not exactly sure what WolframAlpha is doing there, but I don't think it's solving the original equation. The second solution does satisfy the original equation.Hi jacks,
According to Wolfram, both solutions of the quadratic equation, \(x^2+x=1\) are solutions of \(\{x^2\}+\{x\} = 1\). That is,
\[x=-\frac{\sqrt{5}+1}{2}\mbox{ and }x=\frac{\sqrt{5}-1}{2}\]
are solutions of \(\{x^2\}+\{x\} = 1\).
Kind Regards,
Sudharaka.
Yes, as chisigma had mentioned there is an ambiguity in the definition, however taking \(\{x\}=x- \lfloor x \rfloor\,\forall x\in\Re\), both solutions satisfy the original equation.The first solution does not satisfy the original equation. I'm not exactly sure what WolframAlpha is doing there, but I don't think it's solving the original equation. The second solution does satisfy the original equation.
[EDIT]: See posts below for more clarification.
Indeed, but just for the curiosity, can you explain how you thought about this.Forgetting for the moment any problems about negative values of $x$, notice that for any positive integer $n$, the positive solution of the equation $x^2+x = n$ will satisfy $\{x^2\} + \{x\} = 1$, unless that solution is an integer.