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Math Help - Solving radical equations

  1. #1
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    Solving radical equations

     \sqrt(k+2) - \sqrt(k-3) = 1
    \sqrt(k+2) = \sqrt(k-3) + 1
    (\sqrt(k+2))^2 = (1 + \sqrt(k-3))^2
    k+2=1+\sqrt(k-3)+\sqrt(k-3)+(k-3)
    k+2=-2+k+2\sqrt(k-3)
    0=2\sqrt(k-3)
    0=4(k-3)
    0=4k-12
    12=4k
    4=k

    but the answer is k=7
    Where did i mess up?
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  2. #2
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    Quote Originally Posted by m0nkeysensei View Post
     \sqrt(k+2) - \sqrt(k-3) = 1
    \sqrt(k+2) = \sqrt(k-3) + 1
    (\sqrt(k+2))^2 = (1 + \sqrt(k-3))^2
    k+2=1+\sqrt(k-3)+\sqrt(k-3)+(k-3)
    k+2=-2+k+2\sqrt(k-3)
    0=2\sqrt(k-3) \, \, \,\, \textcolor{red}{this \, step \, is \, incorrect \, \, \, \, 4 = 2\sqrt{k-3}}
    0=4(k-3)
    0=4k-12
    12=4k
    4=k

    but the answer is k=7
    Where did i mess up?
    ...
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