A Lagrange Mulipliers Problem My problem is this: Find the min and max values of $f(x,y)=x^2+3xy+y^2$ on the domain $(x-1)^2+y^2=1$. 

I used lagrange multipliers to find that the $y$ coordinate satisfies $f\left(x\right)=8y+(46y^2)/3-16y^3-24y^4=0$ for any critical point $(x,y)$, and $x=2y^2+(2/3)y$
Using Wolfram Alpha, I found that this has roots (aprox.),

$y=-0.97110$

$y=-0.45311$

$y=0.75755$

$y=0$

Wolfram alpha tells me that the function is optimized at the first and third points. However, it is unable to get them in exact form that is not horrendous(multiple radicals involving i). But my professor is expecting an exact answer-and certainly not the exact answer I have. 


Am I missing something?

$f(x,y,\lambda) = x^2 + y^2 + xy + \lambda (x^2 - 2x + y^2)\\
\frac {\partial f}{\partial x} = 2(1+\lambda) x + y - 2\lambda = 0\\
\frac {\partial f}{\partial y} = x + 2(1+\lambda) y = 0\\
\frac {\partial f}{\partial \lambda} = x^2 - 2x + y^2 = 0\\
$ 
$x,y,\lambda = 0$ is one solution (and is the min)
$2\lambda = -\frac {x+2y}{y}\\ 
2(1+\lambda) = -\frac{x}{y}$
$-\frac {x^2}{y} + y + \frac {x+2y}{y} = 0\\
-x^2 + y^2 + x+2y = 0\\
x^2 + y^2 - 2x = 0\\
2y^2 -x + 2y = 0\\
x = \sqrt {1-y^2}+1\\
2y^2 +2y-\sqrt{1-y^2} -1 = 0\\
2y^2 +2y-1 = \sqrt{1-y^2}\\
4y^4 + 8y^3  - 4y + 1 = 1-y^2\\
4y^4 + 8y^3 +y^2 - 4y = 0\\
y(4y^3 + 8y^2 + y - 4)= 0\\
$
And now we need the cubic formula:
$y = \sqrt[3]{\frac {31}{108}- \sqrt{(\frac {31}{108})^2-(\frac {13}{36})^3}}+\sqrt[3]{\frac {31}{108}+ \sqrt{(\frac {31}{108})^2-(\frac {13}{36})^3}}-\frac 23\approx 0.57638\\
x = \sqrt{1-y^2}+1 \approx 1.81718\\
f(x,y) = 4.68175$
