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Now, if I draw the following:
$3x+y=3$
$2x^2-y^2=-1$
With Wolfram I get the following graph.
enter image description here

But if I draw the following functions that I have solved for Y
$y=\sqrt{1+2x^2}$
$y=3-3x$
I seem to be loosing some information
enter image description here

Is all this simply a consequence of the following math:
$y^2=1+2x^2$
$y=\sqrt{1+2x^2}$

Squaring could cause this, but is that what I have done here? The reason for my post is that when doing the calculations by hand I still find the two intersections. But it is not visible in the graphs of the very same functions that on paper gives me what I want.

Algific
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    You're right. To complete things, plot the negative square root as well. – J. M. ain't a mathematician Oct 04 '11 at 22:43
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    Note, for example, that $(0,-1)$ is on the graph of $y^2=1+2x^2$ but not on the graph of $y=\sqrt{1+2x^2}$. When you say you get both intersections by hand, surely that's because somewhere along the way you "square both sides" thereby re-introducing the other branch of the hyperbola. – Gerry Myerson Oct 04 '11 at 22:56
  • Remember that $\sqrt{y^2} = |y|$, not $y$. So your first function should not be $y=\sqrt{1+2x^2}$, but rather $|y|=\sqrt{1+2x^2}$. – Arturo Magidin Nov 28 '11 at 04:32

1 Answers1

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The "missing information" comes from improperly taking the square root. The actual equation B, solved for y, is: $$ y \ =\ \pm\ \sqrt{1 \ +\ 2x \ ^2\ } $$ This will give you the second part of the graph that you were previously missing.

Adam Hammes
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