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How can we know how many solutions an equation has?

  1. Feb 8, 2010 #1
    I have this equation:

    [tex]\frac{x+2}{x^2+3}-0.3[/tex]

    I don't want to solve it, but I have drawn it's graph on a graph program, and there is a line that is little below the x-axis; and when I scroll to the left of the graph, it seems that at more negative values the line is getting higher and higher.

    My question is that how can I know whether it will touch the x-axis or not. In other words, how can I know how many solutions there are to this function?
     
  2. jcsd
  3. Feb 8, 2010 #2
    To say that the function touches the x-axis is another way of stating that for some x-value the resulting value will be 0 (i.e. the y-value is 0). Thus what you want to do is determine whether:
    [tex]\frac{x+2}{x^2+3}-0.3 =0[/tex]
    Consider the function:
    [tex]f(x) = \frac{10x+20}{x^2+3} - 3[/tex]
    This is just your equation scaled by a factor 10 to make the number easier to work with. Clearly scaling won't change the number of intersections with the x-axis. Now note,
    [tex]f(-2) = -3 < 0[/tex]
    [tex]f(0) = 11/3 > 0[/tex]
    [tex]f(8) = -101/67 < 0[/tex]
    and since f is continuous it must intersect the x-axis between -2 and 0 (when going from negative to positive), and again between 0 and 8 (when going from positive to negative).

    To see that these are the only solutions just rewrite f(x) as:
    [tex]f(x) =\frac{10x+20}{x^2+3} - 3\frac{x^2+3}{x^2+3} = \frac{-3x^2 + 10x + 11}{x^2+3}[/tex]
    so f(x) =0 precisely when
    [tex]-3x^2+10x+11=0[/tex]
    but a quadratic function intersects the x-axis at most twice, and we have found two intersections so precisely twice. Alternatively you could just have found this quadratic function and computed the discriminant and noted that it was positive.
     
  4. Feb 8, 2010 #3
    I understand that a quadratic function intersects the x-axis at most twice, but what the denominator of [tex]\frac{-3x^2 + 10x + 11}{x^2+3}?[/tex] Surely that denominator changes the degree of the equation and hence the number of solutions?
     
  5. Feb 8, 2010 #4
    Actually it doesn't change the number of solutions because a/b = 0 if and only if a=0 (to see this just multiply by b to get a=0b=0). Thus if we have,
    [tex]h(x) = \frac{f(x)}{g(x)} \qquad g(x) \not= 0[/tex]
    Then h(x)=0 if and only if f(x)=0.
    EDIT: From these observations it follows that the set of roots of h and f are equal, and therefore they have the same number of roots.
     
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