# fractional part Eq.

#### jacks

##### Well-known member
Calculate $x$ in $\{x^2\}+\{x\} = 1$

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

#### Sudharaka

##### Well-known member
MHB Math Helper
Calculate $x$ in $\{x^2\}+\{x\} = 1$

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

#### Ackbach

##### Indicium Physicus
Staff member
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.
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.

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
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.

#### Sudharaka

##### Well-known member
MHB Math Helper
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.
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.

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.
Indeed, but just for the curiosity, can you explain how you thought about this.

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.