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Solve for x in terms of y (Quadratic formula)

  1. Sep 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Use the quadratic formula to solve the equation for (a) x in terms of y and (b) y in terms of x.

    2. Relevant equations
    4x^2 - 4xy + 1 - y^2 = 0

    3. The attempt at a solution
    I am not really sure where to start at all. If I could just figure out the values for a, b, and c of the quadratic formula then the rest would be simple (for me). Do you take the values from the current equation like this?

    a = 4, b = -4y, c = 1 - y^2

    I feel as if I am over analyzing this problem. If anyone could point me to a start that would be greatly appreciated. There are no examples from my textbook that pertain to this exact type of problem.
  2. jcsd
  3. Sep 12, 2010 #2


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    Gold Member

    Yes, you got it.
  4. Sep 12, 2010 #3
    Thank you for the quick reply. I did run into one more issue with this problem. After plugging in the values into the quadratic formula, I run into this step:

    (4y +– sqrt(32y^2 - 16)) / 8

    There are numbers inside the sqrt that I could pull out: 32 and -16. My next planned step it to re-arrange the values inside the sqrt as follows:

    1) sqrt(32y^2 - 16) = sqrt((16)(2y^2) - (1)(16))
    2) (16) * sqrt(2y^2 - 1)

    This doesn't seem right though. I keep looking for an example, definition, rule, or law to use in this scenario but can't seem to think of one.

    EDIT: Actually, it seems right now that I factored correctly.

    1) sqrt(32y^2 - 16) = sqrt((16)(2y^2 - 1))
    2) (16) * sqrt(2y^2 - 1)

    Then the whole formula will look something like this:

    (4y +- 4 * sqrt(2y^2 - 1)) / 8

    I can then reduce the fraction to (y +- sqrt(2y^2 - 1)) / 2

    I think I've got it! This problem has been stopping me for a while now.
  5. Sep 20, 2010 #4
    Correct! Now, you've solved half the problem (you've solved for x in terms of y) ... you still need to solve for y in terms of x
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