Graphing (x^2)y + 3y^2 = 5x + 9 or the like?

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Homework Statement



Hey everyone. I have a mathematics exam coming up, and I am always used to checking my answers using an alternative method. We have differentiation questions and i normally check them by graphing the equation on my calc and then graphing my calculated differential equation and see if the slope of the original function at any x-value (my calculator can calculate slopes) is the same as the y-value for my differential equation.

My problem is, sometimes there are equations such as these: x2y + 3y2 = 5x + 9 which i can't graph because they have the y^2 and another y which isn't squared. My question is, does anybody know a way around this?

Homework Equations



x2y + 3y2 = 5x + 9

The Attempt at a Solution



I have no idea, i looked online but i didn't know what to google, so all the things that came up were just irrelevant.

Any help would be appreciated.

Thank you.
 
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Solve it as a quadratic equation for y.

ehild
 
How do you do that? It has an x in it... Can you show me how? Thanks
 
The solution(s) of the quadratic equation will be functions of x.

3 y^2+(x^2)y-(5x+9)=0

y_1=\frac{-(x^2)+\sqrt{x^4+12(5x+9)}}{6}

y_2=\frac{-(x^2)-\sqrt{x^4+12(5x+9)}}{6}

ehild
 
nousername said:
My question is, does anybody know a way around this?

[Any help would be appreciated.

Thank you.


Yeah, I gotta' way around this but you may not like my answer. It is however I feel the best approach to solving your dilemma: time to step into the 21st century and do away with that calculator and begin using Mathematica then just ContourPlot it:

Code:
ContourPlot[x^2 y + 3 y^2 == 5 x + 9, {x, -15, 15}, {y, -15, 15}]

Now I understand you need the algebra practice solving them manually but eventually you run into one you can't like:

x^5 y^4 + 3 y^3 x == 5 x^3 y^3 + 9


and that one too can be plotted nicely via ContourPlot
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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