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Question about roots

  1. Dec 12, 2008 #1
    I'm not sure I need to post the whole question, but if i'm having difficulty I will. Basically I'm not sure what the root they are asking me to find is. I have a curve of y=2x2(6-x) and the line y=-3x+36 and I've plotted both of these on a graph. I've shown that where the line and the curve meet 2x3-12x2-3x+36=0 (as the question asked), and now its asking me to "find from your graph approximate values for two positive roots of this equation."

    Now, I thought that roots are where it corsses the Y axis and x=0, but my common sense is telling me it wants me to read off the two values on my graph where the line intercepts the curve. So, its either where the curve meets the X axis (0,0) and (6,0), or the two intercepts between the lines, which my graph implies are (1.9 , 30) and (5.75 , 10.95), but I'm confused about which ones would be considered to two roots???

    thanks.
     
  2. jcsd
  3. Dec 12, 2008 #2

    Mentallic

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    Homework Helper

    The roots of an equation - often also called the zeros - are the points where y=0 satisfies the equation. Thus, where the graph crosses the x-axis. The intercepts of the 2 equations are not the roots of that cubic you provided. I would say go with the (0,0) and (6,0) :smile:
     
  4. Dec 12, 2008 #3

    Mark44

    Staff: Mentor

    The line whose equation is y = -3x + 36, and the parabola whose equation is y = 2x^2(6 - x), meet at the point or points for which the y values for the line and parabola are equal. That is, where -3x + 36 = 2x^2(6 - x).
    Or in other words, where 2x^3 - 12x^2 - 3x + 36 = 0. If you were to graph y = 2x^3 - 12x^2 - 3x + 36, its x-intercepts would be exactly the same as those for the intersection points of the line and the parabola.

    A strategy you might employ is to look for rational roots of the cubic equation above. Any rational number p/q that is a root of this equation must be such that p is a divisor of 36 and q is a divisor of 2. So the possible rational roots are {+/-1, +/-1/2, +/-2, +/-3, +/-3/2, +/-4, +/-6, +/-9, +/-9/2}. This list of 18 numbers probably seems long, but if none of these numbers works, there are no rational roots, so then you can try to find real, irrational roots, which is harder to do.
     
  5. Dec 12, 2008 #4

    epenguin

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    Gold Member

    Where the curve of y=2x2(6-x) and the line y=-3x+36 meet is where they meet so that you can read the x values of that straight off what you plotted already, no? In fact your 1.9 and 5.7 are the values of x at the intersection.

    f(x) = g(x) where the graphs of f and g meet. [here f(x) is 2x2(6-x), g(x) = -3x+36 ]

    f(x) - g(x) = 0 is the same thing as this last equation and that f-g is what your 2x3-12x2-3x+36 is. I.e. your x values are also roots of the equation 2x3-12x2-3x+36=0. They are where the curve y = f(x) - g(x) crosses the x axis. But you do not need to draw this curve when you already got the answer and you were not asked for the y values of the intersections. (They are in fact f(1.9) or, same thing, g(1.9) and f(5.7) or g(5.7) ). All approximate becase you only got and only can get from graphs the roots approximately.

    I am seeing that often not posting the whole question makes it more difficult to help as it is the misunderstanding of the question that causes the problem.
    I also made a mistake in previous version of this post, now edited, I hope it hasn't confused you.
     
    Last edited: Dec 12, 2008
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