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Finding Points of Intersection by Substitution

  1. Jul 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Find any points of intersection of the graphs by the method of substitution.


    2. Relevant equations

    3. The attempt at a solution

    From the second equation I can solve for y:


    Plug it into the first equation and simplify...


    Multiply both sides by 2 to get rid of the fraction and get the radicals over to one side, square both sides


    simplify and you get


    I can find x=0 and that Is correct, but the other real root should be -3. I can graph the polynomial (x3-4x2-x+60) on my calculator and see that it has a root of -3 but how can I do it by hand? Any other way besides the rational roots test?
  2. jcsd
  3. Jul 19, 2010 #2


    Staff: Mentor

    That should be [itex]\pm[/itex].

    Also, it might be simpler to solve for one variable, say y, in the first equation.
  4. Jul 20, 2010 #3


    User Avatar
    Science Advisor

    x(y+1)- 2y+ 3= 0

    x(y+1)= 2y- 3

    Solve that for x and avoid square roots.

  5. Jul 20, 2010 #4
    Ok, so.


    You can sub that into the other equation and get

    [tex]y(4y^3-8y^2-y+6)=0[/tex] Using rational roots test you can find that the root of (4y^3-8y^2-y+6) is 3/2 and of course y= 0 as well.


    Is this what you had in mind? It is easier than having to do the test with all the factors of 60...
  6. Jul 20, 2010 #5


    Staff: Mentor

    Yes, that's what both HallsOfIvy and I had in mind.
  7. Jul 20, 2010 #6


    User Avatar
    Homework Helper

    I'm getting -30 inside the parentheses instead of +6. And a plus 8y2 instead of a minus.

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