1. Determine all pairs $(p, q)$ of real numbers such that the equation $x^{2}+p x+q=0$ has a solution $v$ in the set of real numbers, and if $t$ is a root of this equation, then $|2 t-15|$ is also a root.
SOLUTION. Let $t, s$ be the real roots of the given quadratic equation. First, consider the case where the quadratic equation has a double (real) root. Thus, $t=s$ and according to the conditions of the problem, $t=|2t-15|$. For $t \geq 15/2$, we get the equation $t=2t-15$, with the solution $t=15$. For $t < 15/2$, we get the equation $t=-(2t-15)$, with the solution $t=5$. The corresponding quadratic equations are $(x-5)^2 = x^2 - 10x + 25 = 0$ and $(x-15)^2 = x^2 - 30x + 225 = 0$.

Now, consider the case where the quadratic equation has two distinct real roots $t, s$. We distinguish three cases.

- $t=|2t-15|$ and simultaneously $s=|2s-15|$. The solutions to both equations (see above) form the pair $\{t, s\} = \{5, 15\}$. The corresponding quadratic equation is $(x-5)(x-15) = x^2 - 20x + 75 = 0$.
- $t=|2s-15|$ and simultaneously $s=|2t-15|$. Solving the four systems of equations

$$
t = \pm(2s-15), \quad s = \pm(2t-15)
$$

(which correspond to different choices of signs) yields the pairs $(s, t)$ equal to $(15, 15)$, $(5, 5)$, $(3, 9)$, and $(9, 3)$, of which only the last two satisfy the original system and the condition $s \neq t$. Additionally, the system of equations $t=|2s-15|$ and $s=|2t-15|$ can also be solved graphically in the plane $Ost$, where we plot both broken lines $t=|2s-15|$ and $s=|2t-15|$ (Fig. 1). The pairs $(3, 9)$ and $(9, 3)$ correspond to the quadratic equation $(x-3)(x-9) = x^2 - 12x + 27 = 0$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7bc0c556de791fba4889g-01.jpg?height=514&width=603&top_left_y=1770&top_left_x=778)

Fig. 1

- $t=|2t-15| = |2s-15|$. As we already know, the equation $t=|2t-15|$ has solutions $t=5$ and $t=15$. For $t=5$, the equation $5=|2s-15|$ yields $s=5$ or $s=10$. For $t=15$, the equation $15=|2s-15|$ yields $s=0$ or $s=15$. Considering the condition $s \neq t$, we get two solutions $(t, s) = (5, 10)$ and $(t, s) = (15, 0)$. These solutions correspond to the quadratic equations $(x-5)(x-10) = x^2 - 15x + 50 = 0$ and $(x-15)x = x^2 - 15x = 0$.

Conclusion: The given problem has six pairs $(p, q)$ of real numbers, namely the pairs $(-10, 25)$, $(-30, 225)$, $(-20, 75)$, $(-12, 27)$, $(-15, 50)$, and $(-15, 0)$.

GUIDING PROBLEM:

Solve the system of equations $y=|x-3|, x=3|y-1|$ in the set of real numbers. Verify the result geometrically. [The system has three solutions: $(x, y) = (3, 0)$, $(x, y) = (6, 3)$, and $(x, y) = \left(\frac{3}{2}, \frac{3}{2}\right)$.]