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Math Help - Solve sqrt{x - 5} - sqrt{x - 8) = 3

  1. #1
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    Solve sqrt{x - 5} - sqrt{x - 8) = 3

    \sqrt{x-5} -\sqrt{x-8} =3

    This eventually works out to an (A+B)^2 = A^2 + 2AB +B^2 thing on the right side so that-

    x-8 + 2(\sqrt{x-8}+3) +3^2

    \sqrt{x-5} = x-5 so that

    x-5= x-8 + 2(\sqrt{x-8}+3) +9

    Combining like terms: -8+9 = 1; and then I subtract 1 and x from both sides to get just a -6 on the left side.


    After calculating all this out I came up with 9, but 9 doesn't work. The book says that the answer is 0 with a slash through it. Does that mean no solution because 9 is the only answer and doesn't check? I also checked it with zero and came up with

    -5 +2i\sqrt{2}=3 That can't be right... right?

    I think this may be the problem: I'm thinking that my error may be in 2(\sqrt{x-8}+3) Does that come out to

    6\sqrt{x-8} which when squared =

    36(x-8) which = 36x-288

    Meanwhile on the left side I have 36, derived from -6^2 so that-

    36=-288 +36x: Add 288 to both sides to get 324=36x

    324 divided by 36x= 9

    I realized I've skipped over some steps for the sake of brevity, so if some clarity is needed I'm happy to provide. Any and all help is greatly appreciated. Thanks.
    Last edited by mr fantastic; July 12th 2011 at 06:02 PM. Reason: Re-titled.
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  2. #2
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    Re: Okay, a quick question...maybe not such a quick question

    Isolate the radicals.
    \sqrt{x-5}=3+\sqrt{x-8} so that x-5=9+6\sqrt{x-8}+(x-8).

    Do it a second time.
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  3. #3
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    Re: Okay, a quick question...maybe not such a quick question

    In order to isolate the radical I have to combine like terms and then move them to the other side. Right? Then, in order to elminate the radical from the left side I have to square it, no? At the same time I have to square the left side too, is that correct? And squaring 6\sqrt{x-8} in order to boot out the radical would seem to get me

    36(x-8) which will be

    36x -288


    Am I spazzing out when subtracting x from both sides?

    The example in the book after getting to the point we're talking about has:

    3x+1=x+4+2\sqrt{x+4}+1

    Then it says to combine like terms so that-

    3x+1=x+5+2\sqrt{x+4} which gives

    2x-4=2\sqrt{x+4}


    Then it says to square both sides

    (2x-4)^2=(2\sqrt{x+4}) which gives

    4x^2 -16x +16=4(x+4) so that

    4x^2 -16x +16=4x+16 which gives

    4x^2 -20x=0

    From there you get x=0 and x=5 which I understand well enough. But where exactly am I going wrong in my OP?

    x-x=0 so that would only leave me with a -6 to work with, no?

    I know it's some small detail I'm missing but... I think I might have just realized it... lemme see.

    No, I didn't realize it. I came up with 36(-x +9)=0. That can't be right. Moving 36 from left to right yields me the same thing but in negative. Both are false solutions but maybe they're the right false solutions?
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  4. #4
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    Re: Okay, a quick question...maybe not such a quick question

    You are not missing anything.
    This equation has no solution.
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  5. #5
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    Re: Solve sqrt{x - 5} - sqrt{x - 8) = 3

    Looking at Plato's post (#2), there's no solution by not making any calculation, end of the story.
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