Yeah sorry I fixed it...

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- Feb 11th 2011, 02:46 AMYuuki
Yeah sorry I fixed it...

- Feb 11th 2011, 03:26 AMHallsofIvy
If you replace "m" with "y" it becomes the conic section $\displaystyle x^2+ xy- 6y^2= 0$ 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). - Feb 11th 2011, 04:15 AMemakarov
I think what HallsofIvy is saying is that the fact that $\displaystyle x^2+mx-6m^2=0$ 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.