Is there a typo? It seems to me that Victor and Solomiya just do the same thing...

Tintarn wrote: Is there a typo? It seems to me that Victor and Solomiya just do the same thing... Sorry, edited, thanks

Cute! MS_Kekas wrote: The board contains $20$ non-constant linear functions, not necessarily distinct. For each pair $(f, g)$ of these functions ($190$ pairs in total), Victor writes on the board a quadratic function $f(x)\cdot g(x) - 2$, and Solomiya writes on the board a quadratic function $f(x)g(x)-1$. Victor calculated that exactly $V$ of his quadratic functions have a root, and Solomiya calculated that exactly $S$ of her quadratic functions have a root. Find the largest possible value of $S-V$. Remarks. A linear function $y = kx+b$ is called non-constant if $k\neq 0$. Proposed by Oleksiy Masalitin The answer is $100$. For construction, take ten copies of the linear function $-x$ and ten copies of the linear function $x+2$. Then if you pick $f=g$, $f^2-1$ and $f^2-2$ both have real roots. But $-x(x+2)-2=-(x^2+2x+2)$ has no real roots yet $-x(x+2)-1=-(x+1)^2$ has a real root. So $S-V=100$ is attainable. Now we show this is indeed the maximum. Suppose $f(x)g(x) = Ax^2+Bx+C$. Then $B^2-4AC \ge 0$ as this has two real roots (possibly the same). So $B^2-4A(C-1)=B^2-4AC+4A \ge 4A>0$ and $B^2-4A(C-2) = B^2-4AC+8A \ge 8A>0$, hence both Victor and Solomiya's polynomials have real roots in this case and these terms cancel in $S-V$. So, $S-V$ is no more than the number $N$ of polynomials $f(x)g(x)$ with $f, g$ chosen from the set, with negative leading coefficient. Suppose $p$ of $20$ linear polynomials have positive coefficients and the remaining $20-p$ have negative coefficients, then $S-V \le N = p(20-p) \le 100$, proving the bound.