For the following equation:

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

For which values of $m$ are the roots of the equation real? Examine the signs of the roots as $m$ varies from $-\infty$ to $+\infty$.
The given equation is

$$
x=\frac{1-m \pm \sqrt{1+2 m-3 m^{2}}}{2 m}
$$

or

$$
x=\frac{1-m \pm \sqrt{(1+3 m)(1-m)}}{2 m}
$$

The roots are real if the expression under the square root is positive; both factors are positive if

$$
1 \geqq m \geqq-\frac{1}{3}
$$

Both factors cannot be negative simultaneously, so the roots are real for those values of \( m \) that satisfy the condition (1). The sign of the roots can only be discussed for the boundary values of \( m \). The product and sum of the roots are

$$
\frac{m-1}{m} \text{ and } \frac{-(m-1)}{m} \text{. }
$$

If \( m=-\frac{1}{3} \), then \( x_{1}=x_{2}=-2 \); if \( m \) is greater than \(-\frac{1}{3}\) but less than 0, then the product is positive, while the sum is negative, so both roots are negative. If \( m=0 \), then \( x_{1}=-1, x_{2}=\infty \); if \( m \) is greater than 0 but less than 1, then the product of the roots is negative, so the roots have opposite signs; if \( m=1 \), then \( x_{1}=x_{2}=0 \); finally, if \( m>1 \), then according to (1), the roots are complex.

Summarizing the above, we see that:

$$
\begin{array}{ll}
\text { if }-\infty \leqq m<-\frac{1}{3} \quad \text { then } & \text { the roots are complex, } \\
m=-\frac{1}{3} & x_{1}=x_{2}=-2, \\
-\frac{1}{3}<m<0, & \text { both roots are negative, } \\
m=0 & x_{1}=-1, x_{2}=\infty, \\
0<m<1, & \text { the roots have opposite signs, } \\
1<m \leqq \infty, & \text { the roots are complex. }
\end{array}
$$

(Csete Ferencz Alberik, Eger.)

The problem was also solved by: Filkorn J., Freibauer E., Kornis Ö., Krausz B., Krisztián Gy., Lukhaub Gy., Obláth R., Pollák N., Sasvári G., Tinyó J., Weisz J.