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Math Help - radicals, URGENT!!

  1. #1
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    Exclamation radicals, URGENT!!

    okay I re-did the problem....so the square root of (x-7) + square root of x = 4.
    so I squared both sides to get x-7+x=16.
    simplify: 2x-7=16
    add 7 to both sides 2x=23
    divide each side by 2 and x=11.5
    did i get it?
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  2. #2
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    Quote Originally Posted by iheartthemusic29 View Post
    okay I re-did the problem....so the square root of (x-7) + square root of x = 4.
    so I squared both sides to get x-7+x=16.
    simplify: 2x-7=16
    add 7 to both sides 2x=23
    divide each side by 2 and x=11.5
    did i get it?
    Hello,

    \sqrt{x-7} + \sqrt{x} = 4 . Square both sides of the equation:

    x-7 + 2 \cdot \sqrt{x-7} \cdot \sqrt{x} + x = 16 . Isolate the square-root on one side of the equation:

    2 \cdot \sqrt{x-7} \cdot \sqrt{x} = 23 - 2x . Square both sides:

    4(x-7) \cdot x = 529 - 92x + 4x^2 . Expand the brackets. Collect all like terms:

    4x^2 - 28x = 529 - 92x + 4x^2 \Longrightarrow 64x = 529 \Longrightarrow x = \frac{529}{64}
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  3. #3
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    Hello, iheartthemusic29!

    \sqrt{x-7} + \sqrt{x} \;=\;4

    I squared both sides to get: .  x-7+x\;=\;16 . . . . Wrong-wrong-wrong!
    Isolate a radical: . \sqrt{x-7} \;=\;4 - \sqrt{x}


    Square both sides . . . carefully!

    . . \left(\sqrt{x-7}\right)^2\;=\;(4 - \sqrt{x})^2\quad\Rightarrow\quad x - 7 \;=\;16 - 8\sqrt{x} + x \quad\Rightarrow\quad8\sqrt{x} \;= \;23


    Square both sides: . \left(8\sqrt{x}\right)^2\;=\;23^2\quad\Rightarrow\  quad64x \;=\;529


    Therefore: . \boxed{x \;=\;\frac{529}{64}}

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