Find the quadratic equation equation of $x_1, x_2$. 
Let $x_1 = 1 + \sqrt 3, x_2 = 1-\sqrt 3$. Find the quadratic equation $ax^2+bx+c = 0$ which $x_1, x_2$ are it's solutions.  

By Vieta's theorem:  
$$x_1\cdot x_2 = \frac{c}{a} \implies c=-2a$$
$$x_1 + x_2 = -\frac{b}{a} \implies b = -2a$$
Therefore, $b=c$
So we have a quadratic equation with the form:
$$-\frac{b}{2}x^2 + bx + b = 0$$
Applying $x_1$ for our equation I get $b=0$. Why? 
The final answer is: $x^2-2x-2$.

The quadratic (not quartic) equation of a real number $x$ referres to the monic quadratic polynomial that annihilates $x$. So, the equation $ax^2+bx+c=0$ should only be verified on $x = x_1$ and $x= x_2$.
As expected, making $x = x_1$ you should attain a zero regardless of $b$, you should do that calculation again.
The correct answer should be the one that makes the polynomial monic, i.e., the term $x^2$ with coefficient 1, so $$\frac{b}{2} = -1 \Leftrightarrow b= -2$$ As you observed in the correct answer.
