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Algebra Difficulty 8.2 Shortlist Prove it Hong Kong

Let $x^2 - p_i x + q_i = 0$ , $i = 1, 2, \dots, 15$ be 15 quadratic equations, where each of the $p_i$ and $q_i$ , $i = 1, 2, \dots, 15$ is one of the numbers $1, 2, \dots, 30$ , with no repetition nor omission. Determine the maximum possible number of real roots among the equations which are of values greater than 20.

Solution

The answer is 10.

Let $a \le b$ be two roots of $x^2 - px + q = 0$ such that $b > 20$ . Note that
$a \le \frac{a+b}{2} = \frac{p}{2} \le 15.$

This shows it is impossible to have both roots greater than 20. Also, we have
$b = \frac{p + \sqrt{p^2 - 4q}}{2} < \frac{p+p}{2} = p.$

This implies $p \ge 21$ , and so $p$ can only be 21, 22, $\dots,$ 30. Thus, at most 10 equations may have a root greater than 20.

Consider the equations $x^2 - (20+k)x + k = 0$ for $k = 1, 2, \dots, 10$ . Note that
$\Delta = (20 + k)^2 - 4k = k^2 + 36k + 400 > 0,$

so the roots are real. The larger root is
$\frac{(20 + k) + \sqrt{k^2 + 36k + 400}}{2} > \frac{(20 + k) + (20 - k)}{2} = 20.$

This shows it is possible to have 10 roots larger than 20.

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