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Math Help - Irrational equation help

  1. #1
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    Irrational equation help

    I need explanation.

    I have to solve irrational equation:
    \sqrt {x^2  - 2x + 1}  + \sqrt {x^2  + 2x + 1}  = 2

    I have tried to solve it like this:
    \sqrt {(x - 1)^2 }  + \sqrt {(x + 1)^2 }  = 2
    x-1+x+1  = 2
    x = 1

    But in the book stands that solution is actually interval:
    - 1 \le x \le 1
    which is true.

    How do I find that interval? Can someone explain me solution.
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  2. #2
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    Quote Originally Posted by DenMac21
    I need explanation.

    I have to solve irrational equation:
    \sqrt {x^2  - 2x + 1}  + \sqrt {x^2  + 2x + 1}  = 2

    I have tried to solve it like this:
    \sqrt {(x - 1)^2 }  + \sqrt {(x + 1)^2 }  = 2
    x-1+x+1  = 2
    x = 1

    But in the book stands that solution is actually interval:
    - 1 \le x \le 1
    which is true.

    How do I find that interval? Can someone explain me solution.
    To show that the book is wrong set x=1/2 and then evaluate:

    \sqrt {x^2  - 2x + 1}  + \sqrt {x^2  + 2x + 1}

    Does it equal 2?

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack
    To show that the book is wrong set x=1/2 and then evaluate:

    \sqrt {x^2  - 2x + 1}  + \sqrt {x^2  + 2x + 1}

    Does it equal 2?

    RonL
    Yes, you are right. But, -1 and 0 are also solutions, so author maybe had in mind only integers (even though it is not mentioned anywhere).
    How to come to those solutions?
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  4. #4
    Forum Admin topsquark's Avatar
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    Start with \sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}=2.

    As you noticed, this is:
    \sqrt{(x-1)^2}+\sqrt{(x+1)^2}=2.

    But when we take the square root, we get \pm the answer, so:
    \pm (x-1) + \pm (x+1) = 2, where the \pm signs are independant.

    Using the "+ +" version we get x=1.
    Using the "+ -" version we get -2=2. Nonsense.
    Using the "- +" version we get 2=2. Nothing new.
    Using the "- -" version we get x=-1.

    I admit that I'm baffled by the x=0 solution. I tried squaring twice to get rid of the radicals (the most general method I know of) and I get (x+1)^2=(x+1)^2 (unless I made a mistake) which doesn't help at all.

    -Dan

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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    Start with \sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}=2.

    As you noticed, this is:
    \sqrt{(x-1)^2}+\sqrt{(x+1)^2}=2.

    But when we take the square root, we get \pm the answer, so:
    \pm (x-1) + \pm (x+1) = 2, where the \pm signs are independant.

    Using the "+ +" version we get x=1.
    Using the "+ -" version we get -2=2. Nonsense.
    Using the "- +" version we get 2=2. Nothing new.
    Using the "- -" version we get x=-1.

    I admit that I'm baffled by the x=0 solution. I tried squaring twice to get rid of the radicals (the most general method I know of) and I get (x+1)^2=(x+1)^2 (unless I made a mistake) which doesn't help at all.

    -Dan

    Except \sqrt{x} by (a common) convention denotes the the
    positive root.

    Also \sqrt{u^2}=|u|, so you should have |x-1| + |x+1| = 2,
    (add \pm and \mp signs to taste if you don't want to use the
    positive root convention).

    RonL
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by CaptainBlack
    Except \sqrt{x} by (a common) convention denotes the the
    positive root.

    Also \sqrt{u^2}=|u|, so you should have |x-1| + |x+1| = 2,
    (add \pm and \mp signs to taste if you don't want to use the
    positive root convention).

    RonL
    Exactly right, and what I've just been scratching out on paper.

    I now recommend not looking at the original equation, but the function:
    f(x)=2-\sqrt{x^2+2x+1}-\sqrt{x^2-2x+1}
    This is the function related to the given equation, so we are looking for the zeros of f(x).

    The simplest solution is to graph it. The zeros are obvious...x = [-1,1] as originally advertised. To explain this, note that (as CaptainBlack noted) the convention is to take \sqrt{x} positive. So, taking a look at f(x):
    f(x)=2- \pm (x+1) - \pm (x-1)
    where we now take the +/- signs to require that the term is positive.

    So, on the interval x=(-infinity,-1):
    f(x)=2- -(x+1) - -(x-1)
    (Since (x+1) is negative on this interval we keep the "-" sign to make it positive...similarly (x-1) is negative and we again keep the "-" sign.)
    and thus f(x)=2x+2.

    On the interval x=(1,infinity):
    f(x)=2- +(x+1) - +(x-1)
    and thus f(x)=-2x+2.

    Finally, on the interval x=[-1,1]:
    f(x)=2- +(x+1) -  -(x-1)
    and thus f(x)=0.

    This is an interesting problem. It gives a strong reminder that we can't always simplify an equation and get the same solution set as we had in the unsimplified equation. It's always a good idea to plug your solution back into the original equation to see if you've got everything. (I'm going to note this problem for when I teach College Algebra again!)

    -Dan
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  7. #7
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    Here is a graph:
    (I realized that people have a lot of difficulty when it come to solving absolute value equations, probably has to do with examaning each interval).
    Thus the solution set is,
    -1\leq x\leq 1
    Notice not a finite number of solutions!
    Attached Thumbnails Attached Thumbnails Irrational equation help-picture5.gif  
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