Results 1 to 7 of 7

Math Help - algebra and roots

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    29

    algebra and roots

    Hi, I need a little help. I got this far on a problem but now am stuck, i know i need to find the root but i just cant get it right

    2/(1-d)^2=1/d^2
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JBswe View Post
    Hi, I need a little help. I got this far on a problem but now am stuck, i know i need to find the root but i just cant get it right

    2/(1-d)^2=1/d^2
    do you know what "roots" are?


    \frac 2{(1 - d)^2} = \frac 1{d^2} ........cross multiply

    \Rightarrow (1 - d)^2 = 2d^2

    now expand the left hand side and continue
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Another attempt

    Rearrange

    \frac{(1-d)^2}{d^2}=2

    and extract the respectively square root.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Sep 2007
    Posts
    29
    ok so then i get <br />
(1 - d)^2 = 2d^2

    and when i extract it i find

    <br />
1-d= 2d^2

    2d^2+d-1=0

    d^2+0.5d-0.5=0

    I think i did somthing wrong here. that would give me a neg root the way i tired it. where did i go wrong?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by JBswe View Post
    ok so then i get <br />
(1 - d)^2 = 2d^2

    and when i extract it i find

    <br />
1-d= 2d^2
    huh?!

    (1 - d)^2 \ne 1 - d
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2007
    Posts
    29
    the whole problem looks like this.

    A Positive charge q is between 2 other postive charges Q and 2Q. the distance between Q and 2Q is 1 meter. There is a point on the line between Q and 2Q where the power in the charge q is zero. Where is that point?

    <--------1 meter----->
    +Q---+q-----------+2Q
    <--d-->
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by JBswe View Post
    the whole problem looks like this.

    A Positive charge q is between 2 other postive charges Q and 2Q. the distance between Q and 2Q is 1 meter. There is a point on the line between Q and 2Q where the power in the charge q is zero. Where is that point?

    <--------1 meter----->
    +Q---+q-----------+2Q
    <--d-->
    The "power" in the charge is zero. What does this mean? Are you looking for the point in space where the force on q is zero?

    If so then the net force on charge q at a position d (measuring a positive direction to the right) is:
    F_q = \frac{\kappa q(Q)}{d^2} - \frac{\kappa q(2Q)}{(1 - d)^2}

    So if the net force is 0:
    \frac{\kappa q(Q)}{d^2} - \frac{\kappa q(2Q)}{(1 - d)^2} = 0

    \frac{1}{d^2} - \frac{2}{(1 - d)^2} = 0

    \frac{1}{d^2} = \frac{2}{(1 - d)^2}<-- Multiply both sides by d^2(1 - d)^2

    (1 - d)^2 = 2d^2

    1 - 2d + d^2 = 2d^2

    d^2 + 2d - 1 = 0<-- Use the quadratic formula:

    d = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2}

    d = \frac{-2 \pm \sqrt{8}}{2}

    d = \frac{-2 \pm 2 \sqrt{2}}{2}

    d = -1 \pm \sqrt{2}

    Now, since \sqrt{2} > 1 we must take the "+" solution (since the "-" solution gives a negative d.) Thus
    d = \sqrt{2} - 1

    (Note also that d is less than 1, which makes sense as that puts it between the charges.)

    For the record, the negative value of d also makes sense: There is a point on the far side of Q where the repulsion of q from Q is equal to the attraction of q to the 2Q charge. But, of course, that isn't what we are specifically looking for.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lie Algebra-Roots and Dynkin basis
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 30th 2010, 07:17 PM
  2. Replies: 0
    Last Post: April 24th 2010, 12:37 AM
  3. Roots & Imaginary Roots
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 4th 2009, 10:24 AM
  4. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 10:24 PM
  5. Algebra determine possible roots
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 1st 2007, 02:42 PM

Search Tags


/mathhelpforum @mathhelpforum