Let $a+b=2 \text{ and } ab = -1.$ Determine $a^{10} + b^{10}$ 
Let $a,b \in \mathbb{R}$ for which $$a+b=2  \text{ and } ab = -1.$$ Determine $a^{10}+b^{10}.$

I tried to approach this the following way:
From $ab=-1$ we get that $a = -\frac{1}{b}$ and substituting this to $a+b=2$ yields in $b^2-2b-1=0$. This quadratic has the solutions $b=  1 \pm \sqrt{2}$
Subsituting now $b$ we have that $a= - \frac{1}{1\pm \sqrt{2}}$
I have the feeling that I'm not going in the right direction here. Is there something I'm not seeing since the quadratic formula seems to yield a bit too much work?

You have $4 = (a+b)^2 = a^2+2ab+b^2=a^2+b^2-2$,
so $a^2+b^2 = 6$.  Square and cube this to get
$$a^4+2a^2b^2+b^4 = 36$$
$$a^4+b^4= 36-2(-1)^2 = 34$$
and 
$$a^6+3a^4b^2+3a^2b^4+b^6 $$ $$ = a^b+b^6+3a^2(-1)^2+3b^2(-1)^2 = 6^3=216$$
$$a^6+b^6 = 216 - 3(a^2+b^2) = 216-3(6) = 198.$$
Multiply these two results together:
$$(a^4+b^4)(a^6+b^6) = 34\cdot 198$$
$$a^{10} +b^{10}+a^6b^4+a^4b^6 = 6732$$
$$a^{10}+b^{10} = 6732 - a^2(-1)^4-b^2(-1)^4$$ $$ =6732 -(a^2+b^2) = 6732-6 = 6726.$$
