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.
 A: 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 $x^2 - 2x - 2$.
A: $$p(x)=ax^2+bx+c$$
given that $x=1+\sqrt 3$ so there must me an $x=1- \sqrt 3$ as in a quadratic polynomial when you get the roots the irrational part is from the discriminant as $a,b,c$ are rational.
$$p(x)=(x-(1+\sqrt 3))(x-(1- \sqrt 3))$$
expanding this we get,
$$p(x)=x^2-2x-2$$
A: Let $x = 1+\sqrt{3} \Rightarrow (x-1)^2 = 3 \Rightarrow x^2-2x-2 = 0 \Rightarrow a = k, b = -2k,c = -2k$, and $p(2) = -2 \Rightarrow 4a+2b+c=-2\Rightarrow 4k - 4k - 2k = -2 \Rightarrow k = 1 \Rightarrow (a,b,c) = (1,-2,-2)$.
