Mathematical and Physical Journal
for High Schools
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Problem C. 1157. (February 2013)

C. 1157. For what value of the real parameter a will the equation ax^2+a^2 x+a= \frac 1a have two equal roots?

(5 pont)

Deadline expired on March 11, 2013.

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

Megoldás. Rendezzük az egyenletet:

\(\displaystyle ax^2+a^2x+a-\frac1a=0.\)

Két egyenlő gyök pontosan akkor van, ha az egyenlet diszkriminánsa 0:

\(\displaystyle 0=a^4-4a(a-1/a)=a^4-4a^2+4=(a^2-2)^2.\)

Ez pedig pontosan akkor teljesül, ha \(\displaystyle a^2=2\), vagyis \(\displaystyle a=\pm\sqrt2\).


262 students sent a solution.
5 points:208 students.
4 points:33 students.
2 points:3 students.
1 point:9 students.
0 point:9 students.

Problems in Mathematics of KöMaL, February 2013