Results 1 to 6 of 6

Math Help - [SOLVED] How to solve this equation?

  1. #1
    Member
    Joined
    May 2009
    Posts
    99

    [SOLVED] How to solve this equation?

    By using the substitution u = \frac{1}{x} + x, show that the equation 6x^4 - 25x^3 + 12x^2 - 25x + 6 = 0 can be expressed as 6u^2 - 25u = 0 . Hence, solve the equation giving your answer correct to two decimal places.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    6u^2-25u=0 --------- (1)

    Substituting u=x+\frac{1}{x} in (1),

    6(x+\frac{1}{x})^2-25(x+\frac{1}{x})=0

    6(x^2+2+\frac{1}{x^2})-25(x+\frac{1}{x})=0

    6(x^4+2x^2+1)-25(x^3+x)=0

    6x^4-25x^3+12x^2-25x+6=0
    Last edited by alexmahone; June 30th 2009 at 06:42 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2009
    Posts
    99
    How about the roots? How to solve the roots? The answers given in the book are 3.91, 0.26, i and -i. How did they got it? Showing steps are very much welcomed. Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    6u^2-25u=0

    u(6u-25)=0

    u=0 or u=\frac{25}{6}

    1. u=0

    x+\frac{1}{x}=0

    x=-\frac{1}{x}

    x^2=-1

    x=\pm i

    2. u=\frac{25}{6}

    x+\frac{1}{x}=\frac{25}{6}

    \frac{x^2+1}{x}=\frac{25}{6}

    6x^2-25x+6=0

    x=\frac{25\pm \sqrt{625-144}}{12}

    x=\frac{25\pm \sqrt{481}}{12}

    x=\frac{25\pm 21.93}{12}

    x=3.91 or 0.26
    Last edited by alexmahone; June 30th 2009 at 06:43 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,868
    Thanks
    746
    Hello, mark1950!

    This requires Olympic-level gymnastics . . .


    By using the substitution u = x + \frac{1}{x} .show that the equation:

    6x^4 - 25x^3 + 12x^2 - 25x + 6 \:=\: 0 can be expressed as 6u^2 - 25u \:=\: 0 .

    Hence, solve the equation, giving your answer correct to two decimal places.

    We have: . (6x^4 + 12x^2 + 6) - (25x^3 + 25x) \:=\:0 \quad\Rightarrow\quad 6(x^4 + 2x + 1) - 25(x^3+x) \:=\:0

    Divide by x^2\!:\;\;6\left(x^2 + 2 + \frac{1}{x^2}\right) - 25\left(x + \frac{1}{x}\right) \:=\:0 \quad\Rightarrow\quad 6\left(x + \frac{1}{x}\right)^2 - 25\left(x + \frac{1}{x}\right) \:=\:0

    Let . u \:=\:x+\frac{1}{x}\!:\quad 6u^2 - 25u \:=\:0 \quad\Rightarrow\quad u(6u-25) \:=\:0

    . . And we have: . u \:=\:0, \;u \:=\:\frac{25}{6}



    Back-substitute:

    u \:=\:0 \quad\Rightarrow\quad x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2 + 1 \:=\:0\quad\Rightarrow\quad x^2 \:=\:-1 . . . no real roots

    u \:=\:\frac{25}{6} \quad\Rightarrow\quad x + \frac{1}{x} \:=\:\frac{25}{6} \quad\Rightarrow\quad 6x^2 + 6 \:=\:25x \quad\Rightarrow\quad 6x^2 - 25x + 6 \:=\:0

    . . Quadratic Formula: . x \;=\;\frac{25 \pm\sqrt{481}}{12} \;\approx\;\begin{Bmatrix}3.91 \\ 0.26\end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2009
    Posts
    99
    Thanks a lot, guys! You really helped!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Help with solve equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 16th 2010, 06:58 PM
  2. [SOLVED] solve the equation
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 4th 2010, 02:22 PM
  3. Replies: 3
    Last Post: July 18th 2009, 01:28 AM
  4. [SOLVED] Solve the equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: January 30th 2009, 06:00 AM
  5. [SOLVED] solve equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 15th 2008, 03:49 PM

Search Tags


/mathhelpforum @mathhelpforum