Results 1 to 3 of 3

Thread: Quadratic Eq.

  1. #1
    Member ~berserk's Avatar
    Joined
    Apr 2008
    Posts
    82

    Quadratic Eq.

    Write the quadratic equation such that the sum of its roots is $\displaystyle -5$ and the product of its roots is $\displaystyle 6$ I know that $\displaystyle sum=-b/a$ and $\displaystyle product=c/a$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    May 2008
    Posts
    186
    Quote Originally Posted by ~berserk View Post
    Write the quadratic equation such that the sum of its roots is $\displaystyle -5$ and the product of its roots is $\displaystyle 6$ I know that $\displaystyle sum=-b/a$ and $\displaystyle product=c/a$
    (x - 2)(x - 3)

    x^2 -5x + 6 = 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    Hi, berserk! The easiest way to solve this problem is probably trial and error, but if you want the general solution with a more systematic approach:

    Let the roots of our quadratic $\displaystyle ax^2 + bx + c$ be $\displaystyle x_0\text{ and }x_1$. We have $\displaystyle x_0 + x_1 = -\frac ba = -5$ and $\displaystyle x_0x_1 = \frac ca = 6$. This gives us the system of linear equations:

    $\displaystyle \left\{\begin{array}{rcr}
    -b/a & = & -5\\
    c/a & = & 6
    \end{array}\right.$

    $\displaystyle \Rightarrow\left\{\begin{array}{rcr}
    a & = & b/5\\
    a & = & c/6
    \end{array}\right.$

    $\displaystyle \Rightarrow\frac b5 = \frac c6\Rightarrow b = \frac{5c}6$

    Now, let $\displaystyle c$ be some arbitrary value $\displaystyle t$. Then we have as our solution:

    $\displaystyle \left.\begin{array}{rcrcr}
    a & = & \frac c6 & = & \frac t6\smallskip\\
    b & = & \frac{5c}6 & = & \frac{5t}6\smallskip\\
    c & = & t
    \end{array}\right\}
    \Rightarrow(a,\;b,\;c) = \left(\frac t6,\;\frac{5t}6,\;t\right)$

    Now just pick a (nonzero) value for $\displaystyle t$! For example, with $\displaystyle t = 6$ we have

    $\displaystyle ax^2 + bx + c = x^2 + 5x + 6 = (x - 2)(x - 3)$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 11th 2010, 05:42 PM
  2. quadratic?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Feb 10th 2010, 05:05 AM
  3. Replies: 10
    Last Post: May 6th 2009, 09:52 AM
  4. Replies: 1
    Last Post: Jun 12th 2008, 09:30 PM
  5. Quadratic
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Apr 13th 2007, 07:15 PM

Search Tags


/mathhelpforum @mathhelpforum