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?
 A: An alternative is to get a recurrence. For $n>0$, $(a^n+b^n)(a+b)=(a^{n+1}+b^{n+1})+ab(a^{n-1}+b^{n-1})$. So, for $c_n=a^n+b^n$, we have $c_0=c_1=2$ and $c_{n+1}=2c_n+c_{n-1}$. Which allows to compute $$\color{blue}{c_{10}=6726}.$$
A: 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:
$$(a^4+b^4)(a^6+b^6) = 34\cdot 198$$
$$a^{10} +b^{10}+a^6b^4+a^4b^6 = 6732$$
$$a^{10}+b^{10} = 6732 - a^2(-1)^4-b^2(-1)^4$$ $$ =6732 -(a^2+b^2) = 6732-6 = 6726.$$
A: Another way:  Following your tack, if you choose $b = 1-\sqrt{2}$ then $a=-1/(1-\sqrt{2}) = 1+\sqrt{2}$ after you rationalize the denominator.  So $a$ and $b$ are conjugates.  
If you expand $(1+\sqrt{2})^{10} + (1-\sqrt{2})^{10}$ with binomial theorem, all the odd terms will cancel out.  The 10th row of Pascal's triangle is:
1 10 45 120 210 252 210 120 45 10 1
Eliminate the odd terms and double the even and you have 
$$a^{10}+b^{10} = 2\left(1 + 45\cdot 2+210\cdot 4 + 210\cdot 8 + 45\cdot 16 + 32\right)  $$
$$=6726.$$
