Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation

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Problem C. 852. (April 2006)

C. 852. Solve the following inequality on the set of real numbers: $x^2-3\sqrt{x^2+3}\le 1$ .

(5 pont)

Deadline expired on May 18, 2006.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás: Értelmezési tartomány: minden valós x.

$x^2+3-3\sqrt{x^2+3}-4\leq0,$

$\left(\sqrt{x^2+3}-4\right)\left(\sqrt{x^2+3}+1\right)\leq0.$

A második tényező minden x-re pozitív, ezért $\sqrt{x^2+3}\leq4$ , azaz x² $\leq$ 13. Vagyis a megoldás: $x\in\left[-\sqrt{13};\sqrt{13}\,\,\right]$ .

Statistics:

243 students sent a solution.

5 points: 114 students.

4 points: 49 students.

3 points: 43 students.

2 points: 23 students.

1 point: 7 students.

0 point: 7 students.

Problems in Mathematics of KöMaL, April 2006

243 students sent a solution.
5 points:	114 students.
4 points:	49 students.
3 points:	43 students.
2 points:	23 students.
1 point:	7 students.
0 point:	7 students.