Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem I. 122. (January 2006)

I. 122. Create a plain TeX file in which the formula for the solution of the quadratic equation is derived. Also investigate the number of real solutions. The beginning of the text should contain the usual KöMaL-header (number of the problem, name, city, class, school, e-mail address). The TeX source code (i122.tex) is to be submitted.

(10 pont)

Deadline expired on February 15, 2006.

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

Megoldás.

 \magnification=\magstep1\nopagenumbers\leftline{\bf I. 122.}\leftline{Mekk Elek 10. o. t.}\leftline{Nekeresd, Ezermesterk\'epz\H o Int\'ezet}\leftline{kaposzta@nekeresd.hu}\vskip1cmA m\'asodfok\'u egyenlet \'altal\'anos alakja: $ax^2+bx+c=0$, ahol$a\ne0$. Osztva $a$-val \'es teljes n\'egyzett\'e alak\'\i tva,$$x^2+{b\over a}x+{c \over a}=0$$$${\left(x+{b\over2a}\right)}^2-{b^2\over4a^2}+{c \over a}=0$$$${\left(x+{b\over2a}\right)}^2={b^2-4ac\over4a^2}$$Legyen $D=b^2-4ac$. Ha $D<0$, akkor nincs megold\'as, mert a baloldalmindig nemnegat\'\i v, a jobboldal pedig negat\'\i v. Ha $D\ge0$,akkor n\'egyzetgy\"ok\"ot vonva,$$x+{b\over2a}=\pm{\sqrt{b^2-4ac}\over2a}$$$$x={-b\pm\sqrt{b^2-4ac}\over2a}$$Ha $D=0$, akkor a k\'et \'ert\'ek megegyezik. \"Osszefoglalva, ha ak\'et megold\'ast $x_1$-gyel \'es $x_2$-vel jel\"olj\"uk, akkor$$x_{1,2}={-b\pm\sqrt{b^2-4ac}\over2a}.$$A $D=b^2-4ac$ kifejez\'es a m\'asodfok\'u egyenlet {\itdiszkrimin\'ansa\/}; az egyenletenek $D<0$ eset\'en nincs val\'osmegold\'asa, $D=0$ eset\'en egy val\'os megold\'as l\'etezik, $D>0$ eset\'en pedig kett\H o.\bye

A TeX forrás letölthető innen: i122.tex

### Statistics:

 12 students sent a solution. 10 points: Czigler András, Véges Márton, Vincze János. 8 points: 2 students. 7 points: 3 students. 5 points: 3 students. 4 points: 1 student.

Problems in Information Technology of KöMaL, January 2006