Results 1 to 4 of 4

Thread: Solve Using the Square Root Method

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    83

    Solve Using the Square Root Method

    $\displaystyle $(2x-1)^{2}=-4$$

    I thought this could only be used if k (the constant) was positive.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10
    $\displaystyle \displaystyle \sqrt{(2x-1)^2}=\sqrt{-4}\Rightarrow 2x-1=\pm 2\mathbf{i}\Rightarrow x=\frac{1\pm 2\mathbf{i}}{2}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    You are right, this does not have any real solutions.

    It does have two complex solutions though...

    $\displaystyle \displaystyle (2x - 1)^2 = -4$

    $\displaystyle \displaystyle 2x - 1 = \pm \sqrt{-4}$

    $\displaystyle \displaystyle 2x - 1 = \pm 2i$

    $\displaystyle \displaystyle 2x = 1 \pm 2i$

    $\displaystyle \displaystyle x = \frac{1}{2} \pm i$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    83
    Yes, that's what I came up with, just wasn't certain it was correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Square Root Method
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 25th 2009, 02:14 PM
  2. Quasi-Newton method with square root matrix
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Jun 12th 2009, 07:46 PM
  3. help me solve this square root equation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 18th 2008, 02:09 AM
  4. how to solve by square root principle
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Dec 6th 2006, 10:41 AM
  5. Replies: 2
    Last Post: Apr 29th 2006, 01:13 AM

Search Tags


/mathhelpforum @mathhelpforum