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Math Help - Radical problem

  1. #1
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    Radical problem

    I can't figure this out for the life of me...
    I just need the fastest solution please!
    √x+10= 5-√3-x
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  2. #2
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    Quote Originally Posted by puzzledwithpolynomials View Post
    I can't figure this out for the life of me...
    I just need the fastest solution please!
    √x+10= 5-√3-x
    Hi

    First show that -10 \leq x \leq 3

    Then square both sides
    x+10 = 25 - 10 \sqrt{3-x} + 3 - x

    Re-arrange
    5 \sqrt{3-x} = 9 - x

    Square both sides again and solve the second degree polynomial equation.
    Do not forget to check if the solutions are effectively inside [-10 ; 3]
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  3. #3
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    Could you explain how you re-arranged it?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by running-gag View Post
    Hi

    First show that -10 \leq x \leq 3

    Then square both sides
    x+10 = 25 - 10 \sqrt{3-x} + 3 - x

    Re-arrange
    5 \sqrt{3-x} = 9 - x

    Square both sides again and solve the second degree polynomial equation.
    Do not forget to check if the solutions are effectively inside [-10 ; 3]
    Quote Originally Posted by puzzledwithpolynomials View Post
    Could you explain how you re-arranged it?
    Hi puzzledwithpolynomials,

    Running-gag assumed you meant this:

    \sqrt{x+10}=5-\sqrt{3-x}

    Is that correct, or is it:

    \sqrt{x}+10=5-\sqrt{3}-x? Can't really tell.

    If running-gag is correct, then the simplification would go this way:

    \sqrt{x+10}=5-\sqrt{3-x}

    Square both sides:

    x+10 = 25 - 10 \sqrt{3-x} + 3 - x

    Bring everything to the left side except for the radical term:

    x+10-25-3+x=-10\sqrt{3-x}

    Combine terms:

    2x-18=-10\sqrt{3-x}

    Divide by -2:

    -x+9=5\sqrt{3-x} \ \ \ \ or \ \ \ \ 9-x=5\sqrt{3-x}

    Square both sides again to eliminate the radical:

    (-x+9)^2=(5\sqrt{3-x})^2

    x^2-18x+81=25(3-x)

    x^2-18x+81=75-25x

    Move all terms to the left, combine terms, and set the left side equal zero.

    x^2+7x+6=0

    (x+6)(x+1)=0

    x=-6 \ \ or \ \ x=-1

    Do these solutions check out if you substitute them in the original radical equation? Are they within the domain that Running-gag set up for you at the beginning?

    Good luck.
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  5. #5
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    Ahh. I see it now. Yes, running-gag was correct. I'm sorry about the notation issue.
    Thanks again.
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