Results 1 to 2 of 2

Math Help - Rearranging more formulae

  1. #1
    Newbie evanator's Avatar
    Joined
    Mar 2010
    From
    Ireland
    Posts
    24

    Unhappy Rearranging more formulae

    Hi all,

    A problem I was working on required me to rearrange these two equations to find y in terms of x:


    (\frac{x}{2})^2 + (\frac{y}{4})^2 = 1

    x^2 + y^2 + 2x + 2y + 1 = 0


    Here is my attempt for the first one:


    (\frac{y}{4})^2 = 1 - (\frac{x}{2})^2

    \frac{y}{4} = \sqrt{1 - (\frac{x}{2})^2}

    y = 4\sqrt{1 - (\frac{x}{2})^2}


    And the second one:


    x^2 + 2x + y^2 + 2y = -1

    (x + 1)^2 + (y + 1)^2 = 1

    (y + 1)^2 = 1 - (x + 1)^2

    y + 1 = \sqrt{1 - (x + 1)^2}

    y = \sqrt{1 - (x + 1)^2} - 1


    The answers in the back of the book are:


    y = \pm 4\sqrt{1 - \frac{x^2}{3}}


    y = \pm \sqrt{1 - (1 - x)^2} - 1


    Could someone please tell me where I am going wrong?

    Regards,

    evanator
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,953
    Thanks
    1629
    Quote Originally Posted by evanator View Post
    Hi all,

    A problem I was working on required me to rearrange these two equations to find y in terms of x:


    (\frac{x}{2})^2 + (\frac{y}{4})^2 = 1

    x^2 + y^2 + 2x + 2y + 1 = 0


    Here is my attempt for the first one:


    (\frac{y}{4})^2 = 1 - (\frac{x}{2})^2

    \frac{y}{4} = \sqrt{1 - (\frac{x}{2})^2}

    y = 4\sqrt{1 - (\frac{x}{2})^2}
    If x= 0, \left(\frac{x}{2}\right)^2+ \left(\frac{y}{4}\right)^2= 1 becomes \left(\frac{y}{4}\right)^2= 1

    If y= 4, then \left(\frac{4}{4}\right)^2= 1^2= 1 so y= 4 satisfies that.

    But if y= -4, then \left(\frac{-4}{4}\right)^2= (-1)^2= 1 so y= -4 also satisfies that. Any time you take a square root to solve a quadratic equation, you need both + and -: " \pm.


    And the second one:


    x^2 + 2x + y^2 + 2y = -1

    (x + 1)^2 + (y + 1)^2 = 1
    Very good!

    (y + 1)^2 = 1 - (x + 1)^2

    y + 1 = \sqrt{1 - (x + 1)^2}
    Again, squaring a positive or negative number can give you the same thing- you need " \pm". This should be
    y+ 1= \pm\sqrt{1- (x+ 1)^2}

    y = \sqrt{1 - (x + 1)^2} - 1


    The answers in the back of the book are:


    y = \pm 4\sqrt{1 - \frac{x^2}{3}}


    y = \pm \sqrt{1 - (1 - x)^2} - 1


    Could someone please tell me where I am going wrong?

    Regards,

    evanator
    In both cases you are taking only the "positive" solution to a quadratic.

    If x^2= 4 then x can be either 2 or -2: x= \pm 2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: July 5th 2010, 06:20 AM
  2. rearranging formulae with fractions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 14th 2009, 11:33 AM
  3. Rearranging formulae
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 8th 2009, 11:58 AM
  4. Rearranging Formulae
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 8th 2008, 10:47 AM
  5. Urgent help with rearranging formulae
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 31st 2008, 01:22 PM

Search Tags


/mathhelpforum @mathhelpforum