Solving an equation with fractional part function

[Priyadarshi Raj]

Homework Statement

If $\{ x \}$ denotes the fractional part of x, then solve:
$\{ x \} + \{ -x \} = x^2 + x -6$

It's provided that there are going to be 4 roots of this equation. And two of them will be integers.

Homework Equations

$0 \lt \{ x \} \lt 1~~\text{if}~~x \not\in I$
$\{ x \} = 0~~\text{if}~~x \in I$

The Attempt at a Solution

CASE 1: When $x \in I$
⇒ $0 + 0 = x^2 +x -6$
⇒ $x = -3, 2$
I got two integer roots here.

CASE 2: When $x \not\in I$
$\text{Let, }~~x=i+f ,~~\text{where}~~i \in I~~\text{and}~~0<f<1$
⇒ $\{ i+f \} + \{ -(i+f) \} = (i+f)^2 + (i+f) - 6$
⇒ $f +f = (i^2 +f^2 +2if) + (i+f) -6$
⇒ $f^2 + (2i-1)f + (i^2 + i -6) = 0$
⇒ $f = \frac{(1-2i) \pm \sqrt{(2i-1)^2 -4(i^2 + i -6)}}{2}$
⇒ $f = \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2}$
Now
$0<f<1$
⇒ $0<\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} >0 ~~ \text{and} ~~ \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $(1-2i) \pm \sqrt{8i^2 -23}>0 ~~ \text{and} ~~ (1-2i) \pm \sqrt{8i^2 -23}<2$
⇒ $\pm \sqrt{8i^2 -23}>2i-1 ~~ \text{and} ~~ \pm \sqrt{8i^2 -23}<2i+1$
Squaring,
⇒ $8i^2 -23>4i^2 -4i +1 ~~ \text{and} ~~ 8i^2 -23<4i^2 +4i +1$
⇒ $4i^2 +4i -24>0 ~~ \text{and} ~~ 4i^2 -4i -24<0$
⇒ $i^2 +i -6>0 ~~ \text{and} ~~ i^2 -i -6<0$
⇒ $(i+3)(i-2)>0 ~~ \text{and} ~~ (i-3)(i+2)<0$
⇒ $(i>2 \text{or} i<-3) ~~ \text{and} ~~ (i>-2 \text{and} i<3)$
⇒ $2<i<3$

but this gives no definite solution.

Please help me.
Thank you.

 

[BvU]

Hi,

Can you explain how $$
\{ i+f \} + \{ -(i+f) \} = f + f \quad ? $$ For with e.g $x = 1.3$ I don't get $0.6$ !

 

[Samy_A]

Homework Statement

If $\{ x \}$ denotes the fractional part of x, then solve:
$\{ x \} + \{ -x \} = x^2 + x -6$

It's provided that there are going to be 4 roots of this equation. And two of them will be integers.

Homework Equations

$0 \lt \{ x \} \lt 1~~\text{if}~~x \not\in I$
$\{ x \} = 0~~\text{if}~~x \in I$

The Attempt at a Solution

CASE 1: When $x \in I$
⇒ $0 + 0 = x^2 +x -6$
⇒ $x = -3, 2$
I got two integer roots here.

CASE 2: When $x \not\in I$
$\text{Let, }~~x=i+f ,~~\text{where}~~i \in I~~\text{and}~~0<f<1$
⇒ $\{ i+f \} + \{ -(i+f) \} = (i+f)^2 + (i+f) - 6$
⇒ $f +f = (i^2 +f^2 +2if) + (i+f) -6$
⇒ $f^2 + (2i-1)f + (i^2 + i -6) = 0$
⇒ $f = \frac{(1-2i) \pm \sqrt{(2i-1)^2 -4(i^2 + i -6)}}{2}$
⇒ $f = \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2}$
Now
$0<f<1$
⇒ $0<\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $\frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} >0 ~~ \text{and} ~~ \frac{(1-2i) \pm \sqrt{8i^2 -23}}{2} <1$
⇒ $(1-2i) \pm \sqrt{8i^2 -23}>0 ~~ \text{and} ~~ (1-2i) \pm \sqrt{8i^2 -23}<2$
⇒ $\pm \sqrt{8i^2 -23}>2i-1 ~~ \text{and} ~~ \pm \sqrt{8i^2 -23}<2i+1$
Squaring,
⇒ $8i^2 -23>4i^2 -4i +1 ~~ \text{and} ~~ 8i^2 -23<4i^2 +4i +1$
⇒ $4i^2 +4i -24>0 ~~ \text{and} ~~ 4i^2 -4i -24<0$
⇒ $i^2 +i -6>0 ~~ \text{and} ~~ i^2 -i -6<0$
⇒ $(i+3)(i-2)>0 ~~ \text{and} ~~ (i-3)(i+2)<0$
⇒ $(i>2 \text{or} i<-3) ~~ \text{and} ~~ (i>-2 \text{and} i<3)$
⇒ $2<i<3$

but this gives no definite solution.

Please help me.
Thank you.

You have to tell us how you define $\{ x \}$ for negative $x$.

 

[Ray Vickson]

Homework Statement

If $\{ x \}$ denotes the fractional part of x, then solve:
$\{ x \} + \{ -x \} = x^2 + x -6$

It's provided that there are going to be 4 roots of this equation. And two of them will be integers.Please help me.
Thank you.

Since, for example, we can write $-1.4$ as either $-1 - 0.4$ or as $- 2 + 0.6$, we could consider $\{-1.4\}$ as being either $-0.4$ or $+0.6$. Which one would you take?

 

[BvU]

Based on the first relevant equation I concluded $0.6$ ...

 

[Priyadarshi Raj]

Sorry, I did a silly mistake there.

My teacher said, the fractional part function is defined as:
$\{ x \} = x - [x]$, where [x] is the greatest integer function.

@Ray Vickson , that makes $\{ -1.4 \} = +0.6$
So,
$\text{When}~~ x \not\in I \\ \{ x \} + \{ -x \} \\ \qquad = (x - [x] ) + (-x - [-x]) \\ \qquad= -[i+f] - [-(i+f)] \\ \qquad= -(i) - (-i-1)~~\text{or}~~ -(-i-1) - (i) \\ \qquad= 1\\ ⇒ 1 = x^2 +x -6 \\ ⇒x^2 +x -7 = 0$
and that gives the other two roots.

Thank you everyone!