Solving an equation with fractional part function

  • #1

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.
 

Answers and Replies

  • #2
BvU
Science Advisor
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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## !
 
  • #3
Samy_A
Science Advisor
Homework Helper
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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##.
 
  • #4
Ray Vickson
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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?
 
  • #5
BvU
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Based on the first relevant equation I concluded ## 0.6 ## ...
 
  • #6
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!
 

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