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Absolute value

  • Thread starter w0lfed
  • Start date
  • #1
8
0

Homework Statement



Find all real values of X

| x - 1 | + | x + 1 | < 1

and

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


Homework Equations





The Attempt at a Solution



some how i am thinking i should just solve em as is and all should be fine but i think im lacking a lil bit of fundemental knowledge - some help would be fantastic
 

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
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For the first one, start by squaring both sides. This is OK because both sides are positive, so you don't need to worry about sign changes. Some basic information that you need is:

1.) [itex]|a|^2=a^2[/itex] for any real valued expression [itex]a[/itex].
2.) [itex]|a||b|=|ab|[/itex] for any real valued expressions [itex]a,b[/itex].

For the second one you need the additional piece of information that, if [itex]a>0[/itex], [itex]|x|=a[/itex] implies that [itex]x=a[/itex] or [itex]x=-a[/itex].
 
  • #3
8
0
thanks for your help. i got the first one,
but for the second one if |a|.|b| = |a.b|
i then went on to say
-> |x-1|.|x+2| = 3
-> |(x-1)(x+2)| = 3
-> |x^2 + x - 2| = 3
-> x^2 + x = 5

have i done this write and in what direction am i meant to go from here?
 
  • #4
2
0
Once you got to |x^2 + x - 2| = 3, you need to take Tom Mattson's advice and form 2 equations using the fact that if a > 0, |x| = a implies that x = a or x = -a. Then solve both equations.

Also, what did you get for the first part as your answer?
 
  • #5
311
1
Since you have already solved the first one, here is an alternative interpretation of that equation.

|x-1| = a is the distance of a point x from point 1. similarly |x+1| = b, is the distance of the same point x from -1.

distance b/w points -1 and 1 is 2. Triangle inequality tells us, a + b >= 2 always. Therefore you can directly conclude that the first equation will have no solution because a + b can never be less than 2.


can you think of a similar interpretation for the second expression?
 

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