I'm sure asking this kinda problem is stupid but somehow I have never seen such problems before.

$2{x}^2 + 3{y}^2 =0$ what is $3x+2y$?

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    $\begingroup$ Are you trying to solve this over the real numbers? $\endgroup$ – vociferous_rutabaga Jul 24 '14 at 14:12
  • $\begingroup$ It's a sum of two squares, so the only way to solve the equation is to put $x=0$ and $y=0$, so $3x+2y=0$. If you are allowed to use complex numbers the result can be different $\endgroup$ – Riccardo.Alestra Jul 24 '14 at 14:13
  • $\begingroup$ @Riccardo.Alestra, that's only true given that the OP seeks solutions over the reals. $\endgroup$ – vociferous_rutabaga Jul 24 '14 at 14:15
  • $\begingroup$ Solution could be imaginary numbers $\endgroup$ – most venerable sir Jul 24 '14 at 14:31
  • $\begingroup$ If it's over the imaginary (complex) numbers $\mathbb C$, there will be infinitely many solutions. Choose any complex number $a$, and because $\mathbb C$ is algebraically closed, the equation $2a^2+3y^2=0$ has two solutions, namely $\pm i \sqrt{2/3}a.$ Plugging this in, you have an infinite solution set $(3\pm i \sqrt{2/3})a$ for $a\in \mathbb C$. $\endgroup$ – vociferous_rutabaga Jul 24 '14 at 14:40

$\forall t \in \mathbb{R}$ it holds that $t^2 \geq 0$ and $t^2 = 0 \Leftrightarrow t = 0$.

So it follows that if either $x$ or $y$ is not zero, then $2x^2 + 3y^2 > 0$.

Therefore $x = y = 0$ so $3x + 2y = 0$


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