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• Feb 11th 2011, 02:46 AM
Yuuki
Yeah sorry I fixed it...
• Feb 11th 2011, 03:26 AM
HallsofIvy
Quote:

Originally Posted by Yuuki
Thank you, emakarov, HallsofIvy, and Ithaka/

I have another question.

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

I don't understand why $\displaystyle x=-3$ in both circumstances ($\displaystyle m=1$ and $\displaystyle =-\frac{3}{2}$)
I know you can say that because the problem says "when $\displaystyle x=-3$", 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 $\displaystyle x^2+ xy- 6y^2= 0$ 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 AM
emakarov
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.
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