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Math Help - Help Math ???'s

  1. #1
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    Help Math ???'s

    solve: square root of 4-2x=1+square root of x-3
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  2. #2
    Senior Member OReilly's Avatar
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    Please confirm that problem is:
    \sqrt {4 - 2x}  = 1 + \sqrt {x - 3}
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  3. #3
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    Yes

    that is the problem
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  4. #4
    Senior Member OReilly's Avatar
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    \begin{array}{l}<br />
 \sqrt {4 - 2x}  = 1 + \sqrt {x - 3}  \\ <br />
 \left( {\sqrt {4 - 2x} } \right)^2  = \left( {1 + \sqrt {x - 3} } \right)^2  \\ <br />
 4 - 2x = 1 + 2\sqrt {x - 3}  + x - 3 \\ <br />
 6 - 3x = 2\sqrt {x - 3}  \\ <br />
 \left( {6 - 3x} \right)^2  = \left( {2\sqrt {x - 3} } \right)^2  \\ <br />
 9x^2  - 36x + 36 = 4x - 12 \\ <br />
 9x^2  - 40x + 48 = 0 \\ <br />

    x_{1/2}  = \frac{{40 \pm \sqrt {1600 - 1728} }}{{18}} = \frac{{40 \pm \sqrt { - 128} }}{{18}}

    x can't be real number since under square root is negative number.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by OReilly
    \begin{array}{l}<br />
 \sqrt {4 - 2x}  = 1 + \sqrt {x - 3}  \\ <br />
 \left( {\sqrt {4 - 2x} } \right)^2  = \left( {1 + \sqrt {x - 3} } \right)^2  \\ <br />
 4 - 2x = 1 + 2\sqrt {x - 3}  + x - 3 \\ <br />
 6 - 3x = 2\sqrt {x - 3}  \\ <br />
 \left( {6 - 3x} \right)^2  = \left( {2\sqrt {x - 3} } \right)^2  \\ <br />
 9x^2  - 36x + 36 = 4x - 12 \\ <br />
 9x^2  - 40x + 48 = 0 \\ <br />

    x_{1/2}  = \frac{{40 \pm \sqrt {1600 - 1728} }}{{18}} = \frac{{40 \pm \sqrt { - 128} }}{{18}}

    x can't be real number since under square root is negative number.
    And, just to make things really ugly, you should plug those answers back into the original to see that they actually do solve the problem. I personally wouldn't do this one by hand (uuuuuugly ) but in general, whenever you square both sides of an equation you could be introducing extra solutions, so you need to check those solutions in the original equation.

    -Dan
    Last edited by topsquark; June 26th 2006 at 03:30 PM. Reason: Bad smiley face!
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