Results 1 to 8 of 8
Like Tree1Thanks
  • 1 Post By SlipEternal

Math Help - sum of roots

  1. #1
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    sum of roots

    x is a real number.
    x+8+x+10+x+12+.........5x-2+5x=578

    How many times bigger is the multiplication of x values that supplies this equation than the sum of the x values?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,580
    Thanks
    1017

    Re: sum of roots

    sorry I can't understand this at all.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    Re: sum of roots

    First we will find the sum of x values that supplies this equation. After that we will find the multiplication of x values that supplies this equation and we will find the difference. Answer is -590 but I couldn't get it.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,904
    Thanks
    764

    Re: sum of roots

    It looks like the equation can be rewritten as $\displaystyle \sum_{k=4}^{2x} (x+2k) = 578$.

    Breaking it down:

    $\displaystyle \begin{align*}\sum_{k=4}^{2x} (x+2k) & = \sum_{k=4}^{2x}x + \sum_{k=4}^{2x}2k \\ & = x\sum_{k=4}^{2x}1 + 2\sum_{k=4}^{2x}k \\ & = x(2x-3) + 2\left(\sum_{k=1}^{2x}k - \sum_{k=1}^3 k\right) \\ & = x(2x-3) + 2\left(\dfrac{2x(2x+1)}{2} - \dfrac{3(4)}{2}\right) \\ & = x(2x-3) + 2x(2x+1)-12 \\ & = 6x^2-x-12 = 578\end{align*}$

    So, you get a quadratic equation, and find $x = \dfrac{1\pm 119}{12}$. So, the sum of $x$-values is $10 - \dfrac{59}{6}= \dfrac{1}{6}$ and the product of them is $10 \left(-\dfrac{59}{6}\right) = -\dfrac{295}{3}$.

    So, $\displaystyle -\dfrac{\dfrac{295}{3}}{\dfrac{1}{6}} = -6\dfrac{295}{3} = -590$
    Last edited by SlipEternal; April 16th 2014 at 10:20 AM.
    Thanks from romsek
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    Re: sum of roots

    many thanks.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: sum of roots

    I'm trying to understand this equation:

    Quote Originally Posted by SlipEternal View Post
    $\displaystyle \sum_{k=4}^{2x} (x+2k) = 578$.
    for the case of x = -59/6. X=59/6 makes the series:

    $ \frac {-59} 6 + 8 +\frac {-59} 6 + 10 + \frac {-59} 6 + 12 + ....$

    It never reaches 5 x -59/6 = -49.1666

    I think the problem is that the formula

    $\displaystyle \sum _{k=1} ^{2x} = \frac {2x(2x+1)} 2 $

    works only for positive integer values of 2x. So if we find x= -59/6 it doesn't hold up. I maintain that there is only one solution to the series, and that is x=10.
    Last edited by ebaines; April 17th 2014 at 07:57 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,904
    Thanks
    764

    Re: sum of roots

    Quote Originally Posted by ebaines View Post
    I'm trying to understand this equation:



    for the case of x = -59/6. X=59/6 makes the series:

    $ \frac {-59} 6 + 8 +\frac {-59} 6 + 10 + \frac {-59} 6 + 12 + ....$

    It never reaches 5 x -59/6 = -49.1666

    I think the problem is that the formula

    $\displaystyle \sum _{k=1} ^{2x} = \frac {2x(2x+1)} 2 $

    works only for positive integer values of 2x. So if we find x= -59/6 it doesn't hold up. I maintain that there is only one solution to the series, and that is x=10.
    x = 10 is the only solution to the sum equation, true, but the question asked about the underlying equation that generated the sum equation, which was the quadratic $6x^2-x-590 = 0$.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: sum of roots

    The queston didn't ask about a quadratic - it asked about a sum of a series, and that quadratic equation only defines the sum for positive integers of x. This is like asking what values of x satisfy x+1=5 and approaching it be squaring both sides, rearranging and getting the quadratic x^2 +2x-24 = 0, and concluding that the two possible values for x are 4 and -6. Oh well, it does appear that your method matched what the person who devides the question was expecting.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 4th 2013, 09:27 PM
  2. Replies: 1
    Last Post: February 22nd 2013, 04:10 AM
  3. polynomial n roots derivative n-1 roots
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 16th 2012, 03:24 AM
  4. Roots & Imaginary Roots
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 4th 2009, 09:24 AM
  5. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum