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Algebra: is this solvable?

  1. Apr 6, 2006 #1
    http://home.comcast.net/~andykovacs/equation.GIF [Broken]

    g is a constant. I need to find theta.

    Is there some trick I can do to cancel out the denominator in the root?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 6, 2006 #2


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    and see what type of equation you get for y.
  4. Apr 6, 2006 #3


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    No, you can't cancel- but you can square. I would first divide the entire equation by [itex]g sin(\theta)[/itex] to get
    [tex]\frac{20}{g sin(\theta)}= \frac{14}{g sin(\theta)}+ \sqrt{\frac{14}{g sin(\theta)}}+ 8.41[/tex]
    I would even write it as
    [tex]\frac{10}{7}\frac{14}{g sin(\theta)}= \frac{14}{g sin(\theta)}+ \sqrt{\frac{14}{g sin(\theta)}}+ 8.41[/tex]
    because then I can let [itex]y= \frac{14}{g sin(\theta)}[/itex] and have
    [tex]\frac{10}{7}y= y+ \sqrt{y}+ 8.41[/tex]
    That gives
    [tex]\frac{3}{7}y- 8.41= \sqrt{y}[/tex]
    Square on both sides:
    [tex](\frac{9}{49}y- 8.41)^2= y[/itex]

    Solve that quadratic equation for y and then solve
    [tex] y= \frac{14}{g sin(\theta)}[/tex]
    for [itex]\theta[/itex].
    Last edited by a moderator: May 2, 2017
  5. Apr 6, 2006 #4
    Do you get y=-11.2432 and y=-1.0611 after solving the quadratic equation? If so then I probably screwed up with the physics.
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