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

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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 $\displaystyle a=\pm\sqrt2$.

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