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Intersection points

  1. Aug 7, 2005 #1
    Hi guys, i'm just wondering is it possible to solve the following using algebra to obtain the points of intersection of the two curves f(x) = 6sqrt(x) and
    g(x) = [(x+5)^2]/36

    I got to the point where i reconized that the inverse of g(x) = 6sqrt(x) - 5 which looks alot like the function f(x), any hints or solutions to this problem?
     
  2. jcsd
  3. Aug 7, 2005 #2

    HallsofIvy

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    Yes, the inverse of g is f(x)- 5. I don't know that that helps a lot in finding points of intersection.

    The only way I see of finding the points of intersection is to set f(x)= g(x), square both sides to get rid of the square root, and solve the resulting fourth degree equation
     
  4. Aug 7, 2005 #3

    TD

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    I don't know if you can get there your way (something with the inverse) but tehcnically, it's solvable since you get a 4th-degree polynomial. It won't be 'fun' though. Of course, watch out for introducing solution when squaring.

    [tex]\frac{{\left( {x + 5} \right)^2 }}
    {{36}} = 6\sqrt x \Leftrightarrow \left( {\frac{{\left( {x + 5} \right)^2 }}
    {{36}}} \right)^2 = \left( {6\sqrt x } \right)^2 \Leftrightarrow \frac{{\left( {x + 5} \right)^4 }}
    {{1296}} - 36x = 0[/tex]

    If you'd want to know, mathematica gives me:

    [tex]\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
    72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} - \hfill \\
    \frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} -
    216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} +
    \frac{279936}
    {{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
    72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}}
    {3}}}}{2}\end{gathered} [/tex]

    and

    [tex]\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
    72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} + \hfill \\
    \frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} -
    216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} +
    \frac{279936}
    {{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} +
    72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}}
    {3}}}}{2}\end{gathered} [/tex]

    which are approx: [itex]x \to 0.013452\, \wedge \,x \to 29.150[/itex]
     
    Last edited: Aug 7, 2005
  5. Aug 7, 2005 #4
    Oh, with the inverse thing, i just remembered doing a question a while ago that involved finding the points of intersection between two inverse functions, it was much easier to computer the intersection between one of the functions and y = x since inverse always intersect along that line.

    How do i go about solving the 4th degree polynomial?
     
  6. Aug 7, 2005 #5

    TD

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    Just as there are formula's for the quadratic and cubic, there also exists one for the 4th-degree, named Ferrari.

    http://mathworld.wolfram.com/QuarticEquation.html
     
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