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Algebra Difficulty 8.2 Shortlist Prove it Hong Kong

Let x2pix+qi=0x^2 - p_i x + q_i = 0, i=1,2,,15i = 1, 2, \dots, 15 be 15 quadratic equations, where each of the pip_i and qiq_i, i=1,2,,15i = 1, 2, \dots, 15 is one of the numbers 1,2,,301, 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 aba \le b be two roots of x2px+q=0x^2 - px + q = 0 such that b>20b > 20. Note that
aa+b2=p215. 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=p+p24q2<p+p2=p. b = \frac{p + \sqrt{p^2 - 4q}}{2} < \frac{p+p}{2} = p.

This implies p21p \ge 21, and so pp can only be 21, 22, ,\dots, 30. Thus, at most 10 equations may have a root greater than 20.

Consider the equations x2(20+k)x+k=0x^2 - (20+k)x + k = 0 for k=1,2,,10k = 1, 2, \dots, 10. Note that
Δ=(20+k)24k=k2+36k+400>0, \Delta = (20 + k)^2 - 4k = k^2 + 36k + 400 > 0,

so the roots are real. The larger root is
(20+k)+k2+36k+4002>(20+k)+(20k)2=20. \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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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.