Results 1 to 3 of 3

Math Help - Product + Sum of the Roots

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    54

    Exclamation Product + Sum of the Roots

    Hey guys

    Solve the equation  8x^4-2x^3-27x^2+6x+9=0 given that the sum of two of its roots is zero.

    I have no idea how to solve this. I have never dealt with a quartic before.

    Please, any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    I admit this is not an easy problem.

    Let f(x) = 8x^4-2x^3-27x^2+6x+9

    Suppose \alpha,\beta are roots such that \alpha+\beta = 0. This means \beta = -\alpha.

    Thus we can write f(x) = q(x)(x-\alpha)(x+\alpha) = q(x)(x^2-\alpha^2).

    If we let q(x) = ax^2+bx+c.

    we have 8x^4-2x^3-27x^2+6x+9 = (ax^2+bx+c)(x^2-\alpha^2)\Rightarrow a = 8, b= -2

    And since -\alpha^2b = 2\alpha^2 = 6 \Rightarrow \alpha^2 = 3
    And since -\alpha^2c = -3c = 9 \Rightarrow c = -3.


    Thus we found f(x) = (8x^2-2x-3)(x^2-3)

    Now you can find the roots of the 2 quadratic factors yourself.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,663
    Thanks
    603
    Hello, Sunyata!

    Solve the equation: f(x) \:=\: 8x^4-2x^3-27x^2+6x+9\:=\:0
    given that the sum of two of its roots is zero.
    Let the two roots be: . a\text{ and }-a.

    Then: . \begin{array}{ccccccc}f(a) = 0\!: & a^4 - 2a^3 - 27a^2 + 6a + 9 &= & 0 & [1] \\<br />
f(\text{-}a) = 0\!: & a^4 + 2a^3 - 27a^2 - 6a + 9 &=& 0 & [2] \end{array}

    Subtract [2] - [1]: . 4a^3 - 12a \:=\:0 \quad\Rightarrow\quad 4a(a^2-3) \:=\:0

    . . and we have three roots: . a \:=\:0,\:\pm\sqrt{3}

    . . but a = 0 is an exraneous root.

    Hence, two of the factors are: . (x - \sqrt{3})(x + \sqrt{3})


    Dividing f(x)\text{ by }x^2-3, we get: . 8x^2 - 2x - 3

    The equation becomes: . (x^2-3)(2x+1)(4x-3) \:=\:0


    Therefore, the roots are: . x \;=\;\pm\sqrt{3},\;-\frac{1}{2},\;\frac{3}{4}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sum and product of roots
    Posted in the Algebra Forum
    Replies: 3
    Last Post: July 12th 2011, 10:26 AM
  2. Replies: 5
    Last Post: February 5th 2011, 06:23 AM
  3. Product & Sum of Roots
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 28th 2008, 05:52 PM
  4. Sum and Product of roots ><
    Posted in the Algebra Forum
    Replies: 16
    Last Post: April 7th 2008, 03:18 AM
  5. Replies: 2
    Last Post: September 1st 2007, 12:22 AM

Search Tags


/mathhelpforum @mathhelpforum