Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-42.jpg?height=389&width=519&top_left_y=84&top_left_x=467)
Answer: 0.5 and 1.5.

Solution. Let the origin of the coordinate system be $O$, and the smaller root be $p$ (then the length of the segment $K O$ is also $p$). According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Also, by the problem, $K L=K M$, so $O L=p+2 p=3 p$, i.e., the larger root is $3 p$ (Fig. 7).

![](https://cdn.mathpix.com/cropped/2024_05_06_d7ec6e60d14a8fdbf750g-42.jpg?height=323&width=507&top_left_y=757&top_left_x=477)

Fig. 7: Solution to problem 10.6

By Vieta's theorem, the product of the roots of a quadratic trinomial is equal to the ratio of the constant term to the leading coefficient. Since the constant term is equal to the value of the trinomial at zero, i.e., the length of the segment $O M$, we get the equation

$$
p \cdot 3 p=\frac{\sqrt{3} p}{2 / \sqrt{3}}
$$

Solving this equation, we get $p=0.5$. Therefore, the roots of the trinomial are $p=0.5$ and $3 p=1.5$.