Thread: Equation involving modulus -- help..

I need help with solving this equation:

|x - 1| = |x| + 1

I know that |x| = x when x >= 0 and |x| = -x when x < 0 but I am not sure how to apply it here.

When |x - 1| >= 0 , i.e. when x > 1, x - 1 = x + 1 => 0

Probably my reasoning is wrong thus far so I need help...

Hi

I can see 2 possibilities to answer
1) The equation is equivalent to |x-1| - |x| = 1
You can build a table like this
--------------------oo------------0-----------1------------+oo
-------|x-1|-------------1-x----------1-x----------x-1
--------|x|----------------x------------x-----------x
----|x-1|-|x|------------1------------1-2x----------1

Now you can easily conclude

2) An alternative way is to square the equation
|x-1| = |x| + 1
|x-1|² = (|x| + 1)²
x²-2x+1 = x²+2|x|+1 because |x|²=x²
-x = |x|

Hello, struck!

Solve: .

I'll do it the Long Way . . .


Since both and can be positive or negative,
. . there are four cases to consider:


[1] Both positive: .
. . .Then we have: . . . . impossible

[2] Positive-negative: .
. . .But this means: . . . . impossible

[3] Negative-positive: .
. . .Then we have: .

[4] Negative-negative: .
. . .Then we have: . . . . always true.


The inequality is satisfied in cases [3] and [4]

The solution is: .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


If we graph the two functions, we can "eyeball" the solution.


The graph of is a , vertex at (0,0). .[1]


The graph of . . is [1], moved one unit to the right.

Code:

             \|
             1*           /
              |\         /
              | \       /
              |  \     /
              |   \   /
              |    \ /
      - - - - + - - * - - - -
              |     1

The graph of . .is [1], moved one unit upward.

Code:

              |
        \     |     /
         \    |    /
          \   |   /
           \  |  /
            \ | /
             \|/
             1*
              |
              |
              |
              |
              |
      - - - - + - - - - - - -
              |

Sketch them on the same coordinate system
. . and we can see the solotion: .