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Math Help - One more equation

  1. #1
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    One more equation

    I'm not sure how to begin attacking this one.

    Given the equation 4x^2 - 4xy + 1 - y^2 = 0

    use the Quadratic Formula to solve for
    (a) x in terms of y (b) y in terms of x.

    Can I get a hint?
    Last edited by earachefl; July 10th 2007 at 02:34 PM. Reason: bad grammar
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  2. #2
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    Quote Originally Posted by earachefl View Post
    I'm not sure how to begin attacking this one.

    Given the equation 4x^2 - 4xy + 1 - y^2 = 0

    use the Quadratic Formula to solve for
    (a) x in terms of y (b) y in terms of x.

    Can I get a hint?
    Hint: 4x^2 +x(-4y)+(1-y^2)=0. So what is a,b,c=?
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  3. #3
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    Hmm...

    I get that I need to somehow isolate the variables on one side of the equation. I just can't seem to figure out how. It seems like every way I attack it, x and y stay mixed up together.

    Are you saying to take
    (-4y) and (1-y^2)and group them together somehow?
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  4. #4
    Member Jonboy's Avatar
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    He's saying identify a b and c.

    Remember a quadratic is in the form (a)x^2\,+\,(b)x\,+\,(c)
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  5. #5
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    That was my first guess as to what he meant, and I did plug those numbers (-1, -4, 1) into the Quadratic Formula, and got the answer -2 \pm \sqrt 5. However, this is nowhere near close to the book's answers of

    y = -2x \pm \sqrt{8x^2 +1}

    or

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

    so I'm no closer in understanding.

    Am I supposed to plug the output of the Quadratic Formula in wherever y appears?
    Last edited by earachefl; July 10th 2007 at 03:46 PM. Reason: idea
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    That was my first guess as to what he meant, and I did plug those numbers (-1, -4, 1) into the Quadratic Formula, and got the answer -2 \pm \sqrt 5. However, this is nowhere near close to the book's answers of

    y = -2x \pm \sqrt{8x^2 +1}

    or

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

    so I'm no closer in understanding.

    Am I supposed to plug the output of the Quadratic Formula in wherever y appears?
    where did -1, -4, 1 come from?

    let's try this hint thing one more time. i will give you a blatant one.

    For (a):

    4x^2 - 4xy + 1 - y^2 = 0

    \Rightarrow 4x^2 + (-4y)x + \left( 1 - y^2 \right) = 0

    Here, a = 4 \mbox { , } b = -4y \mbox { , } c = 1 - y^2



    For (b):

    -y^2 + (-4x)y + 4x^2 + 1 = 0

    Here, a = -1 \mbox { , } b = -4x \mbox { , } c = 4x^2 + 1


    Can you continue now?
    Last edited by Jhevon; July 10th 2007 at 11:12 PM.
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  7. #7
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    Thanks to all for the hints. My textbook hasn't given any example of this kind of problem, but just dumped the question in our laps regardless.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by earachefl View Post
    Thanks to all for the hints. My textbook hasn't given any example of this kind of problem, but just dumped the question in our laps regardless.
    did you get the correct solutions?
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