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Thread: Solve THIS!

  1. #1
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    Solve THIS!

    Here is a nice problem which gets many people. Not because it is tricky nor difficult.
    Solve for $\displaystyle x$
    $\displaystyle |x|+|x+1|=1$.
    I think it gets people because they think you can solve it with basic algebraic moves.
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  2. #2
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    x=0, x=-1

    You actually can solve it algebraically.

    Assume x$\displaystyle \geq$0, |x|+|x+1|=2x+1=1 => 2x=0 => x=0.

    Assume x<0, |x|+|x+1|=-x+(-x-1)=-2x-1=1=>-2x=2 => x=-1.
    Last edited by Treadstone 71; Jan 9th 2006 at 05:58 PM.
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  3. #3
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    Quote Originally Posted by Treadstone 71
    x=0, x=-1

    You actually can solve it algebraically.

    Assume x$\displaystyle \geq$0, |x|+|x+1|=2x+1=1 => 2x=0 => x=0.

    Assume x<0, |x|+|x+1|=-x+(-x-1)=-2x-1=1=>-2x=2 => x=-1.
    HA HA got you!! What about $\displaystyle x=-1/2$?
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  4. #4
    TD!
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    It can be done algebraically.

    |x| + |x+1| = 1
    (|x|+|x+1|) = 1 (both sides are non-negative so squaring is allowed)
    2|x(x + 1)| + 2x + 2x + 1 = 1
    2|x(x + 1)| = -2x-2x
    |x(x + 1)| = -x(x+1)
    (|x(x + 1)|) = (-x(x+1)) (*)
    (x(x+1)) = (x(x+1))

    The last line is trivially true for all x, but there's an extra condition in (*).
    You can only square if both sides have the same sign, the LHS is non-negative so the RHS has to be non-negative either. This gives the condition:

    (x(x+1)) ≥ 0

    With solution, after basic algebra: -1 ≤ x ≤ 0
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  5. #5
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    You didn't specify to solve for ALL x, but to find a solution for the equation. I found two solutions.
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  6. #6
    TD!
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    'Solve for x' is generally used when you're asked to find all x which satisfy the equation or inequality.
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  7. #7
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    Quote Originally Posted by Treadstone 71
    You didn't specify to solve for ALL x, but to find a solution for the equation. I found two solutions.
    Come on, Treadstone stop manipulating me! You knew well that you need to solve for all x. You just made a mistake and it happens. I also made many mistakes on these forums read through them and you will find them.

    Anyways, I happen to like this equation because it solutions are not finite. Good job, TD!
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  8. #8
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    $\displaystyle |x+1|+|x|=1$
    The method I used is this:

    If $\displaystyle x\geq0$ then $\displaystyle x+1\geq0$.
    Thus, $\displaystyle x+1+x=1$ thus, $\displaystyle x=0$ one solution.

    If $\displaystyle x<-1$ thus,$\displaystyle x+1<0$ then $\displaystyle x<0$. Thus, $\displaystyle -x-1-x=-1$ thus, $\displaystyle x=0$ but $\displaystyle x<-1$ thus no solutions.

    If $\displaystyle -1\leq x<0$ thus, $\displaystyle x+1-x=1$ but this is always true! Thus, $\displaystyle -1\leq x<0$ are all solutions.

    Now take the union of all the solution sets $\displaystyle -1\leq x\leq 0$

    Another, non-rigorous, way of solving this problem is with a graph. Graph $\displaystyle y=1-|x|$ and graph $\displaystyle y=|x+1|$ and see where the graphs intersect. You will see the graphs coincide on $\displaystyle -1\leq x\leq 0$
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  9. #9
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    Quote Originally Posted by TD!
    'Solve for x' is generally used when you're asked to find all x which satisfy the equation or inequality.
    True, true
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