I am root!

$S = 1 + \sqrt{3} + 1 - \sqrt{3} = 2$

$P = (1+\sqrt{3})(1 - \sqrt{3}) = -2$

Vieta!

$x^2 -2x -2 = 0 \implies a+b+c = \boxed{-3}$

Let the quadratic equation be $aX^2 + bX+ c$

Let the two roots of this equation be $X=r_1$ and $X=r_2$ .

This can be written as: $X-r1= 0$ and $X-r2= 0$

Multiplying both sides: $(X-r1)(X-r2)=0$

Multiplying out the brackets: $X^2 -r_2X- r_1X + r_1r_2= 0$

Simplified quadratic equation: $X^2- (r_1+r_2) X + (r_1r_2)$

Where, r_1= 1+ \sqrt{3} and \(r_2= 1- \sqrt{3}

Therefore:

[(r 1 + r 2 )= (1+ \sqrt{3}) + (1- root(3))= 2\ $r_1\times r_2)=(1+ \sqrt{3})*(1- \sqrt{3}) = 1- 3= -2$

Substituting (r_1+ r_2) and \( (r_1r_2) , the quadratic equation is: $X^2- (2)X+ (-2)=X^2-2X-2$

Comparing this with the quadratic equation at the top gives: $a= 1, b= -2 , c=-2$

Therefore, $a+b+c = 1 - 2- 2=\boxed{ -3}$

Assuming he meant a monic polynomial. That is $a=1$ and the quadratic is of the form $x^2+bx+c=0$

We can use Vieta's to calculate $b$ and $c$ .

$b=-(1+\sqrt3+1-\sqrt3)=-2\, \, \&\, c=(1+\sqrt3)(1-\sqrt3)=1-3=-2$

$\therefore a+b+c=1-2-2=\boxed{-3}$