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Given the following polynomial $q(x,y)=(2x+3y)^2-1$. How would I show that it has roots with any large $x,y$?

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What does arbitrary large coordinates mean? –  Git Gud Feb 5 '13 at 17:28
    
What happens if we choose $(x,y)$ such that $2x+3y=\pm1$? –  Tapu Feb 5 '13 at 17:35
    
By the way all the roots lies on this line. –  Tapu Feb 5 '13 at 17:36
    
@GitGud values of x and y. –  Blue Pony Inc. Feb 5 '13 at 17:51
    
@BluePonyInc. I still don't understand. Hopefully someone who understands the question will provide an answer. –  Git Gud Feb 5 '13 at 17:53

1 Answer 1

I don't speak analyst slang, so I might have misunderstood the question.

Note that $$\begin{align} \displaystyle \{(x,y)\in \mathbb{R}^2\colon q(x,y)=0\}&=\{(x,y)\in \mathbb{R}^2\colon(2x+3y)^2=1\}\\&=\{(x,y)\in \mathbb{R}^2\colon 2x+3y=\pm 1\}\\&=\left\{(x,y)\in \mathbb{R}^2:y=\frac{\pm 1-2x}{3}\right\}.\end{align}$$

As $x$ gets larger, $y$ will get smaller as it will be negative.

Therefore you can't get arbitrarily large roots for the equation.

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