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Thread: Rearranging more formulae

  1. #1
    Newbie evanator's Avatar
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    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:


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

    $\displaystyle x^2 + y^2 + 2x + 2y + 1 = 0$


    Here is my attempt for the first one:


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

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

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


    And the second one:


    $\displaystyle x^2 + 2x + y^2 + 2y = -1$

    $\displaystyle (x + 1)^2 + (y + 1)^2 = 1$

    $\displaystyle (y + 1)^2 = 1 - (x + 1)^2$

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

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


    The answers in the back of the book are:


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


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


    Could someone please tell me where I am going wrong?

    Regards,

    evanator
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  2. #2
    MHF Contributor

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    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:


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

    $\displaystyle x^2 + y^2 + 2x + 2y + 1 = 0$


    Here is my attempt for the first one:


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

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

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

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

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


    And the second one:


    $\displaystyle x^2 + 2x + y^2 + 2y = -1$

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

    $\displaystyle (y + 1)^2 = 1 - (x + 1)^2$

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

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


    The answers in the back of the book are:


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


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


    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 $\displaystyle x^2= 4$ then x can be either 2 or -2: $\displaystyle x= \pm 2$.
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