- Thread starter
- #1

- Thread starter jacks
- Start date

- Thread starter
- #1

- Feb 5, 2012

- 1,621

Hi jacks,Calculate $x$ in $\{x^2\}+\{x\} = 1$

where $\{x\} = $ fractional part of $x$

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.

- Admin
- #3

- Jan 26, 2012

- 4,192

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.

[EDIT]: See posts below for more clarification.

Last edited:

- Feb 13, 2012

- 1,704

http://mathworld.wolfram.com/FractionalPart.html

Kind regards

$\chi$ $\sigma$

- Moderator
- #5

- Feb 7, 2012

- 2,705

- Feb 5, 2012

- 1,621

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.unlessthat solution is an integer.

If we take, \(\{x\}=x- \lfloor x \rfloor\,\forall x\in\Re\) as the definition of \(\{x\}\), I think that all the roots(except integers) of the equations \(x^2+x = n\) are solutions of the original equation. That is,

\[x=\frac{-1\pm\sqrt{4\,n+1}}{2}\mbox{ where }n\in\mathbb{Z}\, \wedge \,x\notin\mathbb{Z}\]

are solutions of \(\{x^2\} + \{x\} = 1\). Furthermore by looking at this, I feel that these seem to be the only solutions of the original equation.

Kind Regards,

Sudharaka.

- Feb 13, 2012

- 1,704

http://www.wolframalpha.com/input/?i=plot+frac(x^2)+++frac(x)+-1+from+0+to+4

In $0<x<4$ there are 16 solutions...

Kind regards

$\chi$ $\sigma$