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Need help with complicated equation

  1. Apr 30, 2016 #1
    Okay, so this equation is coming from a physics problem, which I essentially already solved (the physics part). The only thing I am stuck on is the algebra (I think) portion of it, so the details of the problem itself do not matter. I know the answer already as well, I just can't get to it myself. The equation is:
    5v12 - 3vo2-2v1vo=0
    For simplicity you can use:
    5x2-3y2-2xy=0
    I am solving for v1 which should equal (-3vo)/5
    If someone could just point me in some direction, that would be great. I have no idea where to start, since I have very little experience with solving for multi-variables without 2 equations.
     
  2. jcsd
  3. Apr 30, 2016 #2

    cnh1995

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    This is a quadratic equation for two variables. Look for the two numbers whose sum is equal to the coefficient of the xy term and whose product is equal to the product of coefficients of the square terms.
     
  4. Apr 30, 2016 #3

    SammyS

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    Item #1: Please use the template which is provided for you when starting a new thread.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    One technique for attacking your problem:

    Divide both sides of the equation by v02, or equivalently by y2 .

    You then have a quadratic equation in (v1/v0) or (x/y) .

    If it's still not clear, replace either of those with some other variable, such as "u".
     
  5. Apr 30, 2016 #4

    Ray Vickson

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    You have a single quadratic equation for the single variable ##v_1##, because your ##v_0## is just some input parameter, not a variable (at least, not in the way you described it). So, just use the familiar quadratic solution formula, and simplify it down as much as possible. You ought to find a second root besides the one ##v_1 = -3v_0/5## that you have given.
     
  6. Apr 30, 2016 #5
    Thank you all for your help. I searched for a while and finally found a method to factoring these types of equations. It has been so long since I've had to solve an equation like this, but it all makes sense now. I used a method called the box method to find the factors which ended up being (5x+3y)(x-y), which gave me the answers x=y, and x=(-3y)/5. I also tried dividing all terms by y2 which gave me the equation in terms of x/y. I then subbed in 'u' for x/y which gave me:
    5u2 - 2u -3 = 0, which I also factored and then re-subbed x/y for 'u' to get the same results.

    One final question to further my learning: How exactly would I use the quadratic formula to solve this? I tried plugging in the 'u' equation and got -3/5, but how would I sub back in (x/y) to get it in terms of y?
     
  7. Apr 30, 2016 #6
    Thank you all for your help. I searched for a while and finally found a method to factoring these types of equations. It has been so long since I've had to solve an equation like this, but it all makes sense now. I used a method called the box method to find the factors which ended up being (5x+3y)(x-y), which gave me the answers x=y, and x=(-3y)/5. I also tried dividing all terms by y2 which gave me the equation in terms of x/y. I then subbed in 'u' for x/y which gave me:
    5u2 - 2u -3 = 0, which I also factored and then re-subbed x/y for 'u' to get the same results.

    One final question to further my learning: How exactly would I use the quadratic formula to solve this? I tried plugging in the 'u' equation and got -3/5, but how would I sub back in (x/y) to get it in terms of y?
     
  8. Apr 30, 2016 #7

    PeroK

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    Let's compare two equations:

    ##x^2 + 2bx - 1 = 0##

    And

    ##x^2 + 2yx - 1 = 0##

    So, what precisely is the difference? The first is a quadratic in ##x## with an arbitrary parameter ##b##. The second is an equation in two "variables" ##x## and ##y##.

    But, algebraically, you can solve both equations by the same method. Just complete the square (or apply the quadratic formula) to get:

    ##x = -b \pm \sqrt{b^2 + 1}##

    And

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

    So, what's the difference? Both are quadratics in ##x##. Isn't it just different notation to use ##b## or ##y##?
     
  9. Apr 30, 2016 #8
    Oh okay. So if I am following correctly:
    ##5x^2 - 2xy - 3y^2##

    In terms of y:
    ##a = 5##
    ##b = -2y##
    ##c = -3y^2##

    ##x = \frac{2y\pm \sqrt{4y^2 - 4(5)(-3y^2)}}{10}##
    ##\Rightarrow x = \frac{2y\pm \sqrt{64y^2}}{10}##
    ##\Rightarrow x = \frac{2y \pm 8y}{10}##
    ##\Rightarrow x = [y, \frac{-3y}{5}##]
     
  10. Apr 30, 2016 #9

    PeroK

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    Yes, algebra works with any letters!
     
  11. Apr 30, 2016 #10
    Thank you very much for your help! I know that it works for any letters, I just wasn't sure if it was legal to include ##y## as part of the coefficients.
     
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