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Math Help - Irrational Equations

  1. #1
    Junior Member
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    Irrational Equations

    "I seem to have reached an odd functional impasse. I am stuck. "

    On both.
    Attached Thumbnails Attached Thumbnails Irrational Equations-equations.gif  
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  2. #2
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    Hello, Logic!

    There are not easy problems.
    They required special treatment, delicate handling.

    Here's the first one . . .


    \sqrt[3]{(x-1)^2} + \sqrt[3]{(x-2)^2} \;=\;\tfrac{5}{2}\sqrt[3]{x-2}\sqrt[3]{x-2}

    We have: . (x-1)^{\frac{2}{3}} + (x-2)^{\frac{2}{3}} \;=\;\tfrac{5}{2}(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}}

    . . . . . . 2(x-1)^{\frac{2}{3}} + 2(x-2)^{\frac{2}{3}} \;=\;5(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}}

    . . . . . . 2(x-1)^{\frac{2}{3}} - 5(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}} + 2(x-2)^{\frac{2}{3}} \;=\;0

    Factor: . \bigg[2(x-1)^{\frac{1}{3}} - (x-2)^{\frac{1}{3}}\bigg]\,\bigg[(x-1)^{\frac{1}{3}} - 2(x-2)^{\frac{1}{3}}\bigg] \;=\;0


    And we have two equations to solve:

    2(x-1)^{\frac{1}{3}} - (x-2)^{\frac{1}{3}} \;=\;0 \quad\Rightarrow\quad 2(x-1)^{\frac{1}{3}} \;=\;(x-2)^{\frac{1}{3}}
    . . Cube both sides: . 8(x-1) \;=\;x-2 \quad\Rightarrow\quad 8x-8 \:=\:x-2 \quad\Rightarrow\quad 7x\:=\:6 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{6}{7}}

    (x-1)^{\frac{1}{3}} - 2(x-2)^{\frac{1}{3}} \:=\:0 \quad\Rightarrow\quad 2(x-2)^{\frac{1}{3}} \:=\:(x-1)^{\frac{1}{3}}
    . . Cube both sides: . 8(x-2) \:=\:x-1 \quad\Rightarrow\quad 8x - 16 \:=\:x-1 \quad\Rightarrow\quad 7x \:=\:15 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{15}{7}}

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  3. #3
    Junior Member
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    Aug 2008
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    Thanks!
    This was extremely helpful.

    However, the second one still greatly troubles me. I have several more equations of the like, but none seem to be... inspirational enough for me to come up with a common method for solving them.
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