Finding value of a quadratic The polynomial 
\begin{equation*}
p(x)= ax^2+bx+c 
\end{equation*}
has $1+\sqrt{3}$ as one of it's roots and also $p(2)=-2$. Is there any way to know the value of $a$, $b$ and $c$? I tried but I can form only $2$ equations how can I figure out value of $3$ unknown with $2$ equations so something must be missing.
Also $a$, $b$ and $c$ are rationals.

Hint: If one of the roots is $1 + \sqrt{3}$ and $a, b, c$ are rational then what must the other root be?

Solution 1:
The other root must then be $1 - \sqrt{3}$.
You have two options now: You can make a system of three linear equations in three variables and simply solve.
Or, you can simply expand $\alpha[x - (1 - \sqrt{3})][x - (1 + \sqrt{3})] = \alpha(x^2 - 2x - 2)$ and use $p(2) = -2$ to get $\alpha = 1$.

Solution 2:
We only need two equations if we know that $a, b, c$ are rational.
$$(1+\sqrt{3})^2 a + (1+\sqrt{3})b + c = 0$$
$$4a + 2\sqrt{3} a + b + \sqrt{3} b + c = 0$$
$a, b, c$ are rational, so we know that $-2a = b$ because the $\sqrt{3}$ terms must cancel each other.
So, $4a + 2b + c = -2$ gives us $c = -2$.
Substituting back into the above we get $4a + b = 2$, so $a = 1$ and $b = -2$.
Thus the quadratic is $x^2 - 2x - 2$.
