10.4. The graph of the quadratic trinomial

$$
y=\frac{2}{\sqrt{3}} x^{2}+b x+c
$$

intersects the positive half-axis of the y-axis at point $M$, and the positive half-axis of the x-axis at points $K$ and $L$ (point $L$ is to the right of point $K$). It turns out that $\angle L K M=120^{\circ}$, and $K L=K M$. Find the roots of the given trinomial.
Solution: Consider the triangle $K O M$ ($O$ - the origin). It is a right triangle, and its angle $K$, being adjacent to angle $L K M$, is $60^{\circ}$. Then $O M = O K \cdot \sqrt{3}$ and $K L = K M = 2 O K$. Let the abscissa of point $K$ be $t > 0$. Then the abscissa of point $L$ is $3 t$. The numbers $t$ and $3 t$ are the roots of the given polynomial, so it has the form $y = a(x - t)(x - 3 t)$. According to the condition, its leading coefficient is $2 / \sqrt{3}$; this is the number $a$. Additionally, its graph passes through the point $M(0 ; t \sqrt{3})$. This means that the equality

$$
t \sqrt{3} = \frac{2}{\sqrt{3}} \cdot (0 - t)(0 - 3 t) \Leftrightarrow 3 t = 6 t^{2} \Leftrightarrow t = \frac{1}{2}
$$

holds since $t \neq 0$. Therefore, the roots of the polynomial are $t = 0.5$ and $3 t = 1.5$.

Answer: $\frac{1}{2}$ and $\frac{3}{2}$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Correctly found the ratio of the roots of the quadratic polynomial, roots not found | 2 points |
| Correct answer and an example of a polynomial where it is realized. The absence of other answers is not proven | 1 point |
| Incorrect solution or its absence | 0 points |