## Task B-2.7.

Determine the value of the real parameter $m$ so that the solutions of the equation

$$
(m x-1) \cdot x=m x-2
$$

represent the lengths of the legs of a right triangle with a hypotenuse of length $\frac{5}{6}$.
## Solution.

The given equation can be written in the form $m x^{2}-(m+1) x+2=0$.

The solutions must be real, so the discriminant of the equation must be greater than or equal to zero,

$$
\begin{aligned}
& (m+1)^{2}-8 m \geqslant 0 \\
& m^{2}-6 m+1 \geqslant 0 \\
& m \in(-\infty, 3-2 \sqrt{2}] \cup[3+2 \sqrt{2}, \infty)
\end{aligned}
$$

If the solutions are the lengths of the legs of a right triangle, then the Pythagorean theorem applies:

$$
\begin{array}{ll}
x_{1}^{2}+x_{2}^{2}=\left(\frac{5}{6}\right)^{2}, & 1 \text { point } \\
\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=\frac{25}{36} . & 1 \text { point }
\end{array}
$$

According to Vieta's formulas, we have

$$
\begin{array}{ll}
x_{1}+x_{2}=-\frac{b}{a}=\frac{m+1}{m} & 1 \text { point } \\
x_{1} x_{2}=\frac{c}{a}=\frac{2}{m} & 1 \text { point }
\end{array}
$$

from which the quadratic equation follows

$$
11 m^{2}-72 m+36
$$

whose solutions are $m_{1}=6$ and $m_{2}=\frac{6}{11}$.

Since $m_{2}$ is not in the interval $(-\infty, 3-2 \sqrt{2}] \cup[3+2 \sqrt{2}, \infty)$, the solution to the problem is only $m_{1}=6$.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
January 4, 2015.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH IN AN APPROPRIATE MANNER.