Results 1 to 5 of 5

Thread: Zeros of this equation

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    119

    Zeros of this equation

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

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

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

    Edit: Opps, I made a typo.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Maybe $\displaystyle 3-2\sqrt{2}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    Quote Originally Posted by youngb11 View Post
    $\displaystyle \displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}$

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

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

    $\displaystyle 4u-2u^2=2u(2-u)$

    the zeros are

    $\displaystyle u=0\Rightarrow\ x=3$

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

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

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

    $\displaystyle x=3\pm2\sqrt{2}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, youngb11!

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


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

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

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

    And we have two equations to solve.


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


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

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

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


    Ah, Archie beat me to it!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    119
    Quote Originally Posted by Archie Meade View Post
    $\displaystyle \displaystyle\ u=(x-3)^{\frac{2}{3}}$

    $\displaystyle 4u-2u^2=2u(2-u)$

    the zeros are

    $\displaystyle u=0\Rightarrow\ x=3$

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

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

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

    $\displaystyle x=3\pm2\sqrt{2}$
    Thanks a lot! I overlooked the 2/3 exponent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to find the zeros of an equation?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Apr 4th 2010, 04:15 PM
  2. Zeros of an Equation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jul 9th 2009, 01:59 PM
  3. Zeros
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Sep 28th 2008, 01:24 PM
  4. Replies: 1
    Last Post: Sep 17th 2008, 06:40 PM
  5. Zeros
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: Oct 30th 2007, 12:39 PM

/mathhelpforum @mathhelpforum