One of the roots of the equation $x^{2}+a x+b=0$ is $1+\sqrt{3}$. Find $a$ and $b$, given that they are rational.

#
Substituting $x=1+\sqrt{3}$ into the equation, we arrive at the equality $(4+a+b)+(a+2) \sqrt{3}=0$. Since $\sqrt{3}$ is an irrational number, such an equality is only possible when $4+a+b=0$ and $a+2=0$.

## Answer

$a=b=-2$.