Results 1 to 9 of 9

Math Help - solving the value for (x)

  1. #1
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    173

    Exclamation solving the value for (x)

    6x^4+7x^3-36x^2-7x+6=0 how can i solve this equation and find out the values for (x).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2008
    From
    UK
    Posts
    484
    Thanks
    65

    Re: solving the value for (x)

    Factorise.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: solving the value for (x)

    Quote Originally Posted by srirahulan View Post
    6x^4+7x^3-36x^2-7x+6=0 how can i solve this equation and find out the values for (x).
    Expression =0 when x=2 so (x-2) is a factor. Do long division to investigate other factors.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    173

    Re: solving the value for (x)

    My question is how do you get (x-2) is a factor.i can't understand about your process.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: solving the value for (x)

    It is a general rule that if a polynomial is zero when x=a then (x-a) is a factor of the polynomial
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: solving the value for (x)

    Quote Originally Posted by srirahulan View Post
    My question is how do you get (x-2) is a factor.i can't understand about your process.
    This can be found with trial and error. We know, from the rational root theorem, that every rational root of a polynomial with integer coefficients must be of the form \textstyle\frac pq, where p is a factor of the constant term and q is a factor of the leading coefficient. The constant term in this polynomial is 6 which has factors 1, 2, 3, and 6, and the leading coefficient is also 6. That means the possible rational roots are

    \pm1, \pm2, \pm3, \pm6, \pm\frac12, \pm\frac13, \pm\frac16, \pm\frac23,\ \mathrm{and}\ \pm\frac32.

    By testing each of these with synthetic division, we find that 2 is a root of the polynomial and hence (x - 2) is a factor (we can also determine that \textstyle-3,-\frac12, and \textstyle\frac13 are roots).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member srirahulan's Avatar
    Joined
    Apr 2012
    From
    Srilanka
    Posts
    173

    Re: solving the value for (x)

    Great Thanks.I can understand about it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jun 2012
    From
    California
    Posts
    28
    Thanks
    10

    Re: solving the value for (x)

     \text{This is a palindromic equation of even degree, meaning it has the form:}

     a_0x^n+a_1x^{n-1}+a_2x^{n-2}+\dots+a_2x^2+a_1x+a_0=0.

     \text{Realizing that}\mspace{7mu} x=0 \mspace{7mu} \text{is not the solution of the equation, thus dividing both side by}\mspace{7mu} x^2\mspace{7mu} \text{we would get:}

     6x^2+7x-36-\frac{7}{x}+\frac{6}{x^2}=0\, \Leftrightarrow\, 6\left(x^2+\frac{1}{x^2}\right)+7\left(x-\frac{1}{x}\right)-36=0
     6\left(x^2+\frac{1}{x^2}-2+2\right)+7\left(x-\frac{1}{x}\right)-36=0
     \Leftrightarrow\, 6\left(x-\frac{1}{x}\right)^2+7\left(x-\frac{1}{x}\right)-24=0
     \text{Let} \mspace{7mu} u=x-\frac{1}{x}\mspace{7mu} \text{we get a quadratic equation in terms of \emph{u}}.
     6u^2+7u-24=0\, \Leftrightarrow\, (3u+8)(2u-3)=0\, \Leftrightarrow\, u=-\frac{8}{3} \mspace{7mu} \lor \mspace{7mu} u=\frac{3}{2}
     \text{When}\mspace{7mu} u=-\frac{8}{3}\, \Rightarrow\, x-\frac{1}{x}=-\frac{8}{3}

     \text{You can finish the rest starting from here. Have fun! This is the general method to solve a palindromic equation of even degree.}

     \text{Some equation might not have rational roots, therefore you would have to resort to this method.}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,790
    Thanks
    1531

    Re: solving the value for (x)

    By the "rational root theorem" we can see that any possible rational number roots must be fractions having numerators and denominators factors of 6. A quick search for integer roots shows that 2 and -3 both satisfy the equation and so x- 2 and x+ 3 are factors. Dividing by the those leaves 6x^2+ x- 1= (3x- 1)(2x+ 1) so that x= 1/3 and x= -1/2 are the other two roots.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help solving y'=e^(x-y)
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: November 17th 2011, 05:41 PM
  2. Solving Log
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 21st 2011, 05:19 PM
  3. [SOLVED] Solving a DE
    Posted in the Differential Equations Forum
    Replies: 6
    Last Post: January 6th 2011, 07:41 AM
  4. Solving for x
    Posted in the Pre-Calculus Forum
    Replies: 12
    Last Post: January 27th 2009, 07:52 PM
  5. Replies: 3
    Last Post: October 11th 2006, 09:15 PM

Search Tags


/mathhelpforum @mathhelpforum