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Math Help - Zeros of this equation

  1. #1
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    Zeros of this equation

    \displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}

    Working it out, I got two x-intercepts of \displaystyle x=3 and \displaystyle x=3+2\sqrt{2}

    However, there is apparently a third root. What am I missing:S?

    Edit: Opps, I made a typo.
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  2. #2
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    Maybe 3-2\sqrt{2}
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  3. #3
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    Quote Originally Posted by youngb11 View Post
    \displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}

    Working it out, I got two x-intercepts of \displaystyle x=3 and \displaystyle x=1+2\sqrt{2}

    However, there is apparently a third root. What am I missing:S?
    \displaystyle\ u=(x-3)^{\frac{2}{3}}

    4u-2u^2=2u(2-u)

    the zeros are

    u=0\Rightarrow\ x=3

    \displaystyle\ u=2\Rightarrow\ (x-3)^{\frac{2}{3}}=2

    \Rightarrow\ (x-3)^2=2^3=8

    \Rightarrow\ x-3=\pm\sqrt{8}=\pm2\sqrt{2}

    x=3\pm2\sqrt{2}
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  4. #4
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    Hello, youngb11!

    You took a square root . . . but it's easy to overlook.


    \displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}

    \text{We have: }\;4(x-3)^{\frac{2}{3}} - 2(x-3)^{\frac{4}{3}} \;=\;0

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

    And we have two equations to solve.


    \text{The first equation is: }\;2(x-3)^{\frac{2}{3}} \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:3}


    \text{The second equation is: }\;2 - (x-3)^{\frac{2}{3}} \:=\: 0 \quad \Rightarrow\quad (x-3)^{\frac{2}{3}} \:=\:2

    \text{Raise both sides to the }\tfrac{3}{2}\text{ power.}
    . . \text{This means }cube\text{ both sides: }\;(x-3)^2 \;=\;8
    . . \text{and take the }square\:root\text{ of both sides: }\;x-3 \;=\;\pm\sqrt{8}

    \text{Hence: }\;\boxed{x \:=\:3 \pm2\sqrt{2}}


    Ah, Archie beat me to it!
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\ u=(x-3)^{\frac{2}{3}}

    4u-2u^2=2u(2-u)

    the zeros are

    u=0\Rightarrow\ x=3

    \displaystyle\ u=2\Rightarrow\ (x-3)^{\frac{2}{3}}=2

    \Rightarrow\ (x-3)^2=2^3=8

    \Rightarrow\ x-3=\pm\sqrt{8}=\pm2\sqrt{2}

    x=3\pm2\sqrt{2}
    Thanks a lot! I overlooked the 2/3 exponent.
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