Results 1 to 6 of 6

Thread: [SOLVED] How to solve this equation?

  1. #1
    Member
    Joined
    May 2009
    Posts
    99

    [SOLVED] How to solve this equation?

    By using the substitution $\displaystyle u = \frac{1}{x} + x$, show that the equation $\displaystyle 6x^4 - 25x^3 + 12x^2 - 25x + 6 = 0 $ can be expressed as $\displaystyle 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
    1,114
    Thanks
    7
    $\displaystyle 6u^2-25u=0$ --------- (1)

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

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

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

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

    $\displaystyle 6x^4-25x^3+12x^2-25x+6=0$
    Last edited by alexmahone; Jun 30th 2009 at 05:42 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
    1,114
    Thanks
    7
    $\displaystyle 6u^2-25u=0$

    $\displaystyle u(6u-25)=0$

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

    1. $\displaystyle u=0$

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

    $\displaystyle x=-\frac{1}{x}$

    $\displaystyle x^2=-1$

    $\displaystyle x=\pm i$

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

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

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

    $\displaystyle 6x^2-25x+6=0$

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

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

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

    $\displaystyle x=3.91$ or $\displaystyle 0.26$
    Last edited by alexmahone; Jun 30th 2009 at 05:43 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, mark1950!

    This requires Olympic-level gymnastics . . .


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

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

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

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

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

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

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



    Back-substitute:

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

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

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

    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: Mar 16th 2010, 05:58 PM
  2. [SOLVED] solve the equation
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Feb 4th 2010, 01:22 PM
  3. Replies: 3
    Last Post: Jul 18th 2009, 12:28 AM
  4. [SOLVED] Solve the equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jan 30th 2009, 05:00 AM
  5. [SOLVED] solve equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 15th 2008, 02:49 PM

Search Tags


/mathhelpforum @mathhelpforum