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Thread: Radical problem.

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

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


    Attempts:

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

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


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


    But, the answer sheet says that x=5. Any help?
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  2. #2
    MHF Contributor

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    $\displaystyle \begin{gathered}
    \sqrt {2x + 6} = 1 + \sqrt {x + 4} \hfill \\
    2x + 6 = 1 + 2\sqrt {x + 4} + x + 4 \hfill \\
    x + 1 = 2\sqrt {x + 4} \hfill \\
    x^2 + 2x + 1 = 4\left( {x + 4} \right) \hfill \\
    \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 $\displaystyle x\ge-3.$

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