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I have the following question that puzzles me: How do I determine the number of non-trivial real solutions to the general equation $ax^2 + bxy + cy^2 = 0$ (up to a scalar)?

My attempt was to fix $y \neq 0$ and then see how many non-complex solution the quadratic equation has for $x$. However, I was hoping that there would be a more elegant way to do this.

Thanks for any help!

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  • $\begingroup$ If you assume $y\neq 0$, set $t = \frac{x}{y}$. Find the number of real zeros of $at^2 + bt + c$ and draw your conclusions. $\endgroup$ – Daniel Fischer♦ Dec 12 '15 at 17:13
  • $\begingroup$ Hint: $ax^2+bxy+cy^2=a(x+(b/2a)y)^2+(c-(b^2/4a))y^2$. (Here I assume $a \neq 0$. You can do the analogous thing assuming $c \neq 0$. If both are zero then the question is quite straightforward.) $\endgroup$ – Ian Dec 12 '15 at 17:14
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HINT: $$ax^2+bxy+cy^2=0$$ $$c(y^2+\frac{b}{c}xy+\frac{a}{c}x^2)=0$$ $$c(y+m_1x)(y+m_2x)=0$$ where $m_1+m_2=-\frac{b}{c}$ and $m_1m_2=\frac{a}{c}$

Hope this helps.

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  • $\begingroup$ If such $m_1,m_2$ exist, that is. $\endgroup$ – Ian Dec 12 '15 at 17:15
  • $\begingroup$ @Ian Yes it's an assumption. $\endgroup$ – SchrodingersCat Dec 12 '15 at 17:17

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