Results 1 to 2 of 2

Thread: Another quadratics!

  1. October 26th 2008, 02:33 PM #1
    Gigabytes
    Gigabytes is offline
    Newbie
    Joined
    Oct 2008
    Posts
    6

    Another quadratics!

    What are the zeros of y-7=4x^2

    and Solve square root of 2x^2 - sqaure root of 2 = 0
    Follow Math Help Forum on Facebook and Google+
  2. October 27th 2008, 05:37 AM #2
    earboth
    earboth is offline
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,741
    Thanks
    90
    Quote Originally Posted by Gigabytes View Post
    [1] What are the zeros of y-7=4x^2

    and [2] Solve square root of 2x^2 - sqaure root of 2 = 0
    to [1].
    If I understand this question correctly then y = 0. Therefore solve for x:

    -7 = 4x^2~\implies~x^2=-\dfrac74~\implies~x=\pm i \cdot \dfrac12 \sqrt{7}

    to [2]:

    \sqrt{2x^2} - \sqrt{2} =0~\implies~\sqrt{2} \cdot \sqrt{x^2} = \sqrt{2} \implies~|x| = 1
    Follow Math Help Forum on Facebook and Google+
« find real solutions for equation | complete squares »

Similar Math Help Forum Discussions

  1. quadratics help
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: February 24th 2009, 03:38 PM
  2. Quadratics
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 8th 2008, 01:39 AM
  3. Quadratics
    Posted in the Algebra Forum
    Replies: 8
    Last Post: November 2nd 2008, 05:58 AM
  4. Help with Quadratics
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 19th 2008, 08:09 PM
  5. Quadratics, Help!
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: June 17th 2008, 04:37 PM

Search Tags


/mathhelpforum @mathhelpforum