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Math Help - Radical Equations Dimensions Problem

  1. #16
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    (\sqrt{x+1} + 2)^2  = (\sqrt{3x+1})^2

    (x+1) + 4\sqrt{x+1} + 4 = 3x+1

    combine like terms on the left side ...

    4\sqrt{x+1} + x + 5 = 3x + 1

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

    divide all terms by 2 ...

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

    square again ...

    4(x+1) = x^2 - 4x + 4

    4x + 4 = x^2 - 4x + 4

    0 = x^2 - 8x

    0 = x(x - 8)

    x = 0 ... makes no sense in the context of the problem

    x = 8
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  2. #17
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    Quote Originally Posted by skeeter View Post
    (\sqrt{x+1} + 2)^2  = (\sqrt{3x+1})^2

    (x+1) + 4\sqrt{x+1} + 4 = 3x+1

    combine like terms on the left side ...

    4\sqrt{x+1} + x + 5 = 3x + 1

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

    divide all terms by 2 ...

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

    square again ...

    4(x+1) = x^2 - 4x + 4

    4x + 4 = x^2 - 4x + 4

    0 = x^2 - 8x

    0 = x(x - 8)

    x = 0 ... makes no sense in the context of the problem

    x = 8
    Alright so in order to get the dimensions I have to input x into the original widths and heights correct?

    Thanks alot for the help man.
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