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Which of these terms is greater ?

$2x-6y+1$ or $1$

if $x^4 + 3y^2=0$

According to the text they are equal ?How is that ?

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  • $\begingroup$ It boils down to comparing $2x-6y$ and $0$, which further reduces to comparing $x$ and $3y$. $\endgroup$
    – Karthik C
    Commented Jul 13, 2012 at 5:36

3 Answers 3

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What number system are $x,y$ in? If $x,y\in\mathbb{R}$, then it is simple, since $x^4$, $3y^2$ $\ge 0$, we must have that $x,y=0$, hence $2x-6y+1=1$.

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But since $x^2$ and $3y^2$ are positive, (assuming $x,y$ real) we have that $x^2+3y^2=0$ forces $x=y=0$ so the text is correct!

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If sum of squares is $0$ then each term of the series is $0$ individually (if they belong to $\Bbb R$) $\implies$ $x=0,y=0$ in your problem which gives $2x-6y+1=1$.

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