There are $11$ quadratic equations on the board, where each coefficient is replaced by a star. Initially, each of them looks like this $$\star x^2 + \star x + \star= 0.$$Two players are playing a game making alternating moves. In one move each ofthem replaces one star with a real nonzero number. The first player tries to make as many equations as possible without roots and the second player tries to make the number of equations without roots as small as possible. What is the maximal number of equations without roots that the first player can achieve if the second player plays to her best? Describe the strategies of both players.
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Tags: quadratics, combinatorics, algebra
26.03.2024 17:53
You misspelled coefficient btw parmenides51 wrote: where each coeffiient is replaced by a star also when you say "without roots", do you mean "without real roots"?
26.03.2024 18:02
any quadratic has always 2 roots in C, so in order for this problem to exist, it means real roots
20.08.2024 19:29
parmenides51 wrote: There are $11$ quadratic equations on the board, where each coefficient is replaced by a star. Initially, each of them looks like this $$\star x^2 + \star x + \star= 0.$$Two players are playing a game making alternating moves. In one move each ofthem replaces one star with a real nonzero number. The first player tries to make as many equations as possible without roots and the second player tries to make the number of equations without roots as small as possible. What is the maximal number of equations without roots that the first player can achieve if the second player plays to her best? Describe the strategies of both players. I’ll called the first player and second player A and B respectively. Consider an equation $ax^2+bx+c=0$. A wants $b^2-4ac<0$ while B wants $b^2-4ac>0$ Suppose $a$ has been replaced by a non-zero number while $a$, $c$ remain stars in an equation, if A is the next player, he choose one of $b$, $c$, $B$ can choose another variable properly to accomplish his goal If B is the next player, he can choose $c$ such that $ac<0$, then no matter what B choose as $b$, $b^2-4ac>0$ So if $a$ has been replaced, B can ensure this equation to be no real root (1) This conclusion is also true when $c$ has been replaced since $a$ and $c$ are symmetric in $b^2-4ac$ Suppose $b$ has been given while $b$, $c$ remain stars If one player replaced $a$ or $c$, the next player can choose another variable properly to accomplish his goal So when $b$ has been replaced, the player who replaced another variable next can’t ensure to accomplish his goal (2) As the best strategy, A will replaced $b$ as much as possible and B will replaced $a$ or $c$ as much as possible In the first $11$ round, A replaced $6$ of $b$ and B replaced $5$ of $a$ or $c$ in the other equations So B might ensure $5$ equation with real roots (applying (1)) And since $2 \nmid 11$, it’s easy to prove that A can ensure another $6$ equation to be non-real root (applying (2)) So the answer is $6$