Results 1 to 6 of 6
Like Tree1Thanks
  • 1 Post By Reckoner

Math Help - quadratic function solution

  1. #1
    Newbie
    Joined
    Jun 2012
    From
    UK
    Posts
    2

    quadratic function solution

    Hi,


    I have a simple question regarding solving a quadratic function
    y = ax^2 + bx + c
    I tried to solve it as follows in general for y=y1 find the roots of the quadratic equation:
    0 = ax^2 + bx + c - y1


    Hence, in more detail...
    for y = y1:
    k = c - y1
    d = b^2 - 4 a k
    assuming d > 0
    x1 = \frac{(-b + sqrt(d))}{2a}
    x2 = \frac{(-b - sqrt(d))}{2a}


    However, this is not producing the desired result. For example, assume the following function


    y(x) = 3x^2 + 2x + 1
    y(3) = 34
    and x(34) following the solution above gives:
    x1=27 and x2=-33
    None of them being 3 as I would expect. Can someone point me in the right direction or a link with info I can look up?


    Thank you in advance.
    Last edited by gatemaze; June 18th 2012 at 06:47 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jun 2012
    From
    France
    Posts
    45
    Thanks
    13

    Re: quadratic function solution

    you have to distinguish between factorization and solving of the equation.

    To solve the equation you have x1 and x2 as you mentioned, this gives you the roots of your equation only. But, the factorization of the the quadratic function is given by
    ax^2+bx+c=a\left(x-\frac{-b+\sqrt{b^2-4ac}}{2a} \right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right).
    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
    75
    Awards
    1

    Re: quadratic function solution

    I'm getting the correct result:

    34 = 3x^2 + 2x + 1

    \Rightarrow0 = 3x^2 + 2x -33

    \Rightarrow x = \frac{-2\pm\sqrt{2^2-4\cdot3(-33)}}{2\cdot3}

    \Rightarrow x = \frac{-2\pm\sqrt{400}}6

    \Rightarrow x = \frac{-2\pm20}6 = \frac{-1\pm10}3

    \Rightarrow x = 3\quad\mathrm{or}\quad x=-\frac{11}3.

    If you are getting something different, I suggest showing your work so that we may locate any errors.
    Thanks from gatemaze
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Jun 2012
    From
    UK
    Posts
    2

    Re: quadratic function solution

    Thanks. After few hours of googling, posting here and reassuring me that it is correct I realised that excel does not like lack of parentheses. Thanks again a lot.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Re: quadratic function solution

    Quote Originally Posted by gatemaze View Post
    Thanks. After few hours of googling, posting here and reassuring me that it is correct I realised that excel does not like lack of parentheses. Thanks again a lot.
    I believe Excel follows the standard order of operations. You'll always need parentheses in situations like that when you're entering an expression in plain text. The horizontal line in a fraction (called a vinculum) acts as a grouping symbol in handwritten expressions, but we don't have that luxury when we just use a / for division.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: quadratic function solution

    To solve ax^2 + bx + c = 0 in general, first divide everything by a:


    x^2 + \frac{b}{a}x + \frac{c}{a} = 0. Complete the square by adding \frac{b^2}{4a^2} to both sides:


    x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{c}{a} = \frac{b^2}{4a^2}


    (x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}


    Take square root of both sides


    x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}


    x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: January 4th 2012, 11:04 PM
  2. Groups; solution to quadratic.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 12th 2010, 06:17 AM
  3. Groups; solution to quadratic.
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: September 11th 2010, 10:50 AM
  4. Quadratic Equation Solution Set
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 4th 2009, 05:25 AM
  5. quadratic solution
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 29th 2009, 05:22 AM

Search Tags


/mathhelpforum @mathhelpforum