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: $x^2-2x-2$.
 A: Is $x_1$ and $x_2$ are the solution than you can write
$$(x_1-x)(x_2-x)=0$$
which is
$$x^2-(x_1+x_2)\cdot x+x_1x_2=0$$
Thus $a=1$, $b=-x_1-x_2$ and $c=x_1x_2$
A: 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.
A: Here's how I would do the "calculate and check" steps:
$$\begin{align}(x-x_1)(x-x_2)&=0\\
x_1&=1+\sqrt 3\\
x_2&=1-\sqrt 3\\
x^2-(x_1+x_2)x+x_1x_2&=0\\
x_1+x_2&=2\\
x_1x_2&=1^2-\sqrt 3^2=-2\\
x^2-2x-2&=0\end{align}$$
Now, taking the form that you created, we have
$$-\frac b2x^2+bx+b = 0$$
Plugging in $x=x_1=1+\sqrt 3$ results in
$$\begin{align}-\frac b2(1^2+2\sqrt 3+3)+b(1+\sqrt 3)+b&=0\tag 1\\
-\frac b2(4+2\sqrt 3)+2b+b\sqrt 3&=0\\
-2b-b\sqrt 3+2b+b\sqrt3 &= 0\\
0 &= 0\end{align}$$
Note that using $x=x_2=1-\sqrt 3$ simply places a couple negative signs so we end up with $(1^2-2\sqrt 3+3)$ and $(1-\sqrt 3)$ in $(1)$.
This check step should always result in $0=0$ when the equation and other calculations are done correctly.
A: From your equation:
$−\frac{b}{2}x^2+bx+b=0$
notice that it's equal to $b*(−\frac{1}{2}x^2+x+1)=0$
Multiplying both sides of the equation by -2 we get:
$b*(x^2-2x-2)=0$
So the solutions to this become either $b=0$
or $x^2-2x-2=0$
Which imply $x=1 \pm \sqrt3$
