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Math Help - solving for x

  1. #1
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    solving for x

    My algebra is rusty:

    I'm solving for x in the following equation:

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

    After several steps (in some of which I square both sides of the equality), I end up with the equality

    x(25x-16) = 0

    At this point I'm afraid to eliminate either term of the product in fear of dividing by 0. So I solve for x. I get x=16/25 , 0.

    Maple says the only solution is x = 16/25.

    Am I supposed to solve this way, but then check the solutions with the original equation in order to compensate for any misleading algebra brought on by me multiplying both sides by a variable?
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  2. #2
    Senior Member furor celtica's Avatar
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    it is always helpful to plug in the values after as some answers can be illogical.
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  3. #3
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    You should always check answers. In this case, 0 is not a valid solution as x= 0 gives the square root of -1.
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  4. #4
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    x^1/2 +(x-(1-x)^1/2)^1/2=1
    =>x-(1-x)^1/2=1+x-2x^1/2
    =>2x^1/2-1=(1-x)^1/2
    =>5x=4x^1/2
    =>x^1/2(5x^1/2-4)=0
    =>x=0,2/sqrt(5)

    wonder what solutions will come if we cosider x=r(cos(theta)+isin(theta))
    ofcourse that would give the actual solution set.
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  5. #5
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    x^1/2 +(x-(1-x)^1/2)^1/2=1
    =>x-(1-x)^1/2=1+x-2x^1/2
    =>2x^1/2-1=(1-x)^1/2
    =>5x=4x^1/2
    =>x^1/2(5x^1/2-4)=0
    =>x=0,2/sqrt(5)

    i wonder what solutions will come if we cosider x=r(cos(theta)+isin(theta))
    ofcourse that would give the actual solution set.
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  6. #6
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    please ignore the 0

    Quote Originally Posted by HallsofIvy View Post
    You should always check answers. In this case, 0 is not a valid solution as x= 0 gives the square root of -1.
    my mistake!!please ignore the zero. also i havenot checked the other answer.
    using eulers formula is worth considering as well
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