sum of CoefficIent

As, $x = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1}$

We get, $x = 3 + 2 \sqrt{2}$

As, $ax^2 - bx + c = 0$ is a Binomial Equation, there have another root, $x = 3 - 2 \sqrt{2}$

So, $(x - 3 - 2 \sqrt{2})(x - 3 + 2 \sqrt{2})=0$

At last, $x^2 - 6x + 1 = 0$

Compairing with, $ax^2 - bx + c = 0$ we get, $a=1, b=6, c=1$

So, $a + b + c = 1 + 6 + 1 = \boxed{8}$

I think, the question shouldn't be for only 10 points... But, what to do? :)

$x=\frac{\sqrt{2}+1}{\sqrt{2}-1}\frac{\sqrt{2}+1}{\sqrt{2}+1} =(\sqrt{2}+1)^2=3+2\sqrt{2}=\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$ .

This gives $b=6a$ and $8=\frac{b^2-4ac}{(2a)^2}=9-\frac{c}{a}$ , meaning that $c=a$ . All in all this gives us:

$ax^2-bx+c=a(x^2-6x+1)$

and $a+b+c=a+6a+a=\boxed{8a}$ . This is valid for any $a\in\mathbb{R}$ , hence as of yet this problem has no single solution. With the (in a way logical) choice of $a=0$ however, we obtain $a+b+c=\boxed{8}$ .