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Math Help - Quadratic Equation Question

  1. #16
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    Yeah sorry I fixed it...
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  2. #17
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    Quote Originally Posted by Yuuki View Post
    Thank you, emakarov, HallsofIvy, and Ithaka/

    I have another question.



    I was able to solve the problem;
    When m=1, x=2 or x=-3
    When m=-\frac{3}{2}, x=\frac{9}{2} or x=-3

    I don't understand why x=-3 in both circumstances ( m=1 and =-\frac{3}{2})
    I know you can say that because the problem says "when x=-3", but I'm not satisfied with that answer.
    I think I may be content if I do the same problem with a different way of approaching it...
    If you replace "m" with "y" it becomes the conic section x^2+ xy- 6y^2= 0 which is a hyperbola having axes of symmetry at an angle about 5 degrees off the vertical. Any horizontal line (value of y or m) crosses it in two different places, giving two different values of x, while any vertical line (value of x) crosses it in two different places, giving two different values for y (or m).
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  3. #18
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    I think what HallsofIvy is saying is that the fact that x^2+mx-6m^2=0 does not imply that x has to be -3 in any way. You can take different values of x and get different values of m or vice versa. The problem arbitrarily says to find m that satisfies the equation when x = -3. You found correctly that m = -1.5 or m = 1.
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