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Math Help - Radical problem.

  1. #1
    Member integral's Avatar
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    Radical problem.

    \sqrt{2x+6}-\sqrt{x+4}=1
    solve for x.


    Attempts:

    (\sqrt{2x+6}-\sqrt{x+4})\sqrt{2x+6}-\sqrt{x+4}=1(\sqrt{2x+6}-\sqrt{x+4})
    2x+6+x+4=\sqrt{2x+6}-\sqrt{x+4}

    2x+6+x+4+\sqrt{x+4}=\sqrt{2x+6}
    2x+6=(3x+10)^2+x+4
    2x+6=9x^2+61x+104
    0=9x^2+59x+98


    x=\frac{-59\pm\sqrt{47}i}{18}


    But, the answer sheet says that x=5. Any help?
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  2. #2
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    \begin{gathered}<br />
  \sqrt {2x + 6}  = 1 + \sqrt {x + 4}  \hfill \\<br />
  2x + 6 = 1 + 2\sqrt {x + 4}  + x + 4 \hfill \\<br />
  x + 1 = 2\sqrt {x + 4}  \hfill \\<br />
  x^2  + 2x + 1 = 4\left( {x + 4} \right) \hfill \\ <br />
\end{gathered}
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    note that if the solution exists, then it must lie on x\ge-3.

    write the inequality as \sqrt{2x+6}=1+\sqrt{x+4}, since both sides are positive, we can square them and get 2x+6=1+2\sqrt{x+4}+x+4 then x+1=2\sqrt{x+4}, in order to square we require also x\ge-1, and of course the previous condition, so we need x\ge-1 in order to make these things work, thus x^2+2x+1=4x+16 then x^2-2x-15=0\implies(x+3)(x-5)=0, we require that x\ge-1, so x=5 is the solution to our problem.
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