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Math Help - Square roots subtraction

  1. #1
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    Square roots subtraction

    I have a problem that says
    √x+8) - √x-4) = 2

    I don't know how to continue with this problem...



    Thank you in advance.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Joaco View Post
    I have a problem that says
    √x+8) - √x-4) = 2

    I don't know how to continue with this problem...
    Thank you in advance.
    Do you mean

    A. \sqrt{x+8} - \sqrt{x-4} = 2 or

    B. \sqrt{x}+8 - (\sqrt{x}-4) = 2 or

    C. \sqrt{x}+8-\sqrt{x}-4 = 2


    If, as I suspect, the correct expression is A then you will need to square both sides

    (x+8) - 2\sqrt{(x+8)(x-4)} + (x-4) = 4

    Simplfying: \sqrt{(x+8)(x-4)} = x

    Square again: (x+8)(x-4) = x^2

    I'll leave you to finish that off


    edit: this equation has only one solution and it is in the domain of the original equation so there can be no extraneous solutions
    Last edited by e^(i*pi); September 12th 2010 at 11:57 AM. Reason: see post
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  3. #3
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    I would have done that slightly differently. From \sqrt{x+ 8}- \sqrt{x- 4}= 2, add \sqrt{x- 4} to both sides to get \sqrt{x+ 8}= \sqrt{x- 4}+ 2. Squaring both sides of that you get x+ 8= x- 4+ 4\sqrt{x- 4}+ 4. You still have the square root form the "cross term" but since you don't have both square roots on the same side, it's a little simpler. That can be reduced, by canceling the "x" terms on both sides and canceling the "-4" and "4", to 8= 4\sqrt{x- 4} which reduces to \sqrt{x- 4}= 2. Now square both sides again to get x- 4= 4.

    Notice that while e^(i*pi)'s equation (x+ 8)(x- 4)= x^2 looks like a quadratic, it is not be cause the [itex]x^2[/itex] terms on each side will cancel.
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    I would have done that slightly differently. From \sqrt{x+ 8}- \sqrt{x- 4}= 2, add \sqrt{x- 4} to both sides to get \sqrt{x+ 8}= \sqrt{x- 4}+ 2. Squaring both sides of that you get x+ 8= x- 4+ 4\sqrt{x- 4}+ 4. You still have the square root form the "cross term" but since you don't have both square roots on the same side, it's a little simpler. That can be reduced, by canceling the "x" terms on both sides and canceling the "-4" and "4", to 8= 4\sqrt{x- 4} which reduces to \sqrt{x- 4}= 2. Now square both sides again to get x- 4= 4.

    Notice that while e^(i*pi)'s equation (x+ 8)(x- 4)= x^2 looks like a quadratic, it is not be cause the [itex]x^2[/itex] terms on each side will cancel.
    That helped me out a lot, I was stuck at the 8= 4\sqrt{x- 4} part until I read your post.
    Thank you a lot.


    Also thank you a lot \sqrt{x+ 8}- \sqrt{x- 4}= 2.
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