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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1Are you trying to solve this over the real numbers? – vociferous_rutabaga Jul 24 '14 at 14:12
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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 – Riccardo.Alestra Jul 24 '14 at 14:13
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@Riccardo.Alestra, that's only true given that the OP seeks solutions over the reals. – vociferous_rutabaga Jul 24 '14 at 14:15
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Solution could be imaginary numbers – most venerable sir Jul 24 '14 at 14:31
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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 $a\in \mathbb C$. – vociferous_rutabaga Jul 24 '14 at 14:40
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$\forall t \in \mathbb{R}$ 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 $3x + 2y = 0$
Darth Geek
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