Find Solutions for |x-1|*|x+1|=0 with Step-by-Step Guide

  • Thread starter Thread starter solar nebula
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around solving the equation |x-1|*|x+1|=0, with participants exploring the implications of absolute values in the context of finding solutions. The original poster expresses uncertainty about their solution process and seeks clarification.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the original poster's method of setting each absolute value expression to zero and question the validity of similar approaches for different equations involving absolute values. There is a focus on understanding the conditions under which certain algebraic manipulations are valid.

Discussion Status

Some participants affirm the correctness of the original poster's solution for the first equation, while others point out flaws in reasoning when applying similar methods to a different equation. The conversation is ongoing, with various interpretations and approaches being explored.

Contextual Notes

Participants note the importance of correctly applying algebraic principles when dealing with absolute values and the implications of setting products equal to zero. There is also a mention of the appropriateness of the thread's placement within the forum.

solar nebula
Messages
14
Reaction score
0

Homework Statement


find all the numbers x for which |x-1|*|x+1|=0

Homework Equations





The Attempt at a Solution


|x-1|*|x+1|=0

|x-1|=0 or |x+1|=0

x-1=0 or x+1=0

x=1 or x=-1

but I think there's a flaw some in my attempt solution...or the whole thing is a flaw..
any help will be much appreciated!
Thanks!
 
Physics news on Phys.org
It's completely correct!
 
micromass said:
It's completely correct!

But when I try to do the same thing for |x-1|*|x+2|=3 which is:

|x-1|*|x+2|=3

|x-1|=3 or |x+2|=3

x-1=3 or x+2=3

x=4 or x=1

But this solution does not make sense...
please help, Thanks!
 
Last edited:
Of course not. You basically did "xy=3 thus x=3 or y=3". This is obviously false.

To solve that, first do the following:

[tex]|x-1||x+1|=3[/tex]

thus

[tex]|(x-1)(x+1)|=3[/tex]

So the absolute value of the number (x-1)(x+1) equal 3. Then what can the number (x-1)(x+1) equal?

BTW I moved this to precalculus, because it obviously does not belong in calculus and beyond. Please post in the correct forum next time.
 
In both cases you are following the procedure
[itex]a*b=c \Rightarrow a=\frac{c}{b}[/itex]
which works as long as [itex]b\neq0[/itex]. When you do it withe [itex]c=0[/itex] then [itex]\frac{c}{b}=0[/itex], but this is true for any [itex]b[/itex] (again, I'm assuming b is not zero). So you just skip right to [itex]a=0[/itex] or [itex]b=0[/itex].

micromass's explanation is probably more intuitive, but I like to think of things algorithmically.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
4
Views
3K
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K