## Task A-2.1. (4 points)

For which values of the real parameter $m$ does the equation

$$
(m-1) x^{2}-2 m x+2=0
$$

have no negative real solutions?
## First solution.

For $m=1$ the observed equation is linear $(-2 x+2=0)$, and its solution $x=1$ is not negative.

For $m \neq 1$, the discriminant of the equation

$$
D=4 m^{2}-4 \cdot 2 \cdot(m-1)=4 m^{2}-8 m+8=(2 m-2)^{2}+4
$$

is positive, so this quadratic equation always has two real solutions, $x_{1}$ and $x_{2}$.

The conditions $x_{1} \geqslant 0, x_{2} \geqslant 0$ are satisfied if and only if $x_{1}+x_{2} \geqslant 0$ and $x_{1} x_{2} \geqslant 0$.

1 point

By applying Vieta's formulas, we get

$$
x_{1}+x_{2}=\frac{2 m}{m-1}, \quad x_{1} x_{2}=\frac{2}{m-1}
$$

Thus, it must hold that $\frac{2 m}{m-1} \geqslant 0, \frac{2}{m-1} \geqslant 0$.

From the first inequality, we get $m \in\langle-\infty, 0] \cup\langle 1, \infty\rangle$, and from the second $m>1$.

Finally, the initial equation has no negative solutions for $m \geqslant 1$.