## Task B-2.2.

Determine the quadratic function whose zeros are $\frac{2 \sqrt{3}}{1-\sqrt{3}}$ and $\frac{2 \sqrt{3}}{1+\sqrt{3}}$ if the point $A(\sqrt{3}, 9)$ lies on the graph of this function.
## Solution.

Let $x_{1}=\frac{2 \sqrt{3}}{1-\sqrt{3}}$ and $x_{2}=\frac{2 \sqrt{3}}{1+\sqrt{3}}$.

By rationalizing the denominators, we get $x_{1}=-3-\sqrt{3}$ and $x_{2}=3-\sqrt{3}$.

The desired function can be written in the form

$$
f(x)=a\left(x-x_{1}\right)\left(x-x_{2}\right), \quad \text { or } \quad f(x)=a\left(x^{2}-\left(x_{1}+x_{2}\right) x+x_{1} x_{2}\right)
$$

After substituting $x_{1}$ and $x_{2}$, we get

$$
f(x)=a\left(x^{2}+2 \sqrt{3} x-6\right)
$$

Since the point $A(\sqrt{3}, 9)$ lies on the graph of this function, we have

$$
9=a\left((\sqrt{3})^{2}+2 \sqrt{3} \cdot \sqrt{3}-6\right)
$$

thus $a=3$. 1 point

The desired quadratic function is $f(x)=3 x^{2}+6 \sqrt{3} x-18$. 1 point

Note: Accept the quadratic function written in factored form $f(x)=$ $3(x+3+\sqrt{3})(x-3+\sqrt{3})$ or vertex form $f(x)=3(x+\sqrt{3})^{2}-27$.