JBMO 2007 problema 1
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Laurian Filip
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JBMO 2007 problema 1
Fie \( a \) un numar real pozitiv astfel incat \( a^3=6(a+1) \). Sa se arate ca ecuatia \( x^2 + ax + a^2 - 6 = 0 \) nu are solutii reale.
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Beniamin Bogosel
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Discriminantul ecuatiei este\( \Delta=24-3a^2 \).
Presupunem ca \( a^2\leq 8 \). Atunci avem
\( 6a+6=a^3\leq 8a \Rightarrow a\geq 3 \Rightarrow a^3\geq 3a^2\Rightarrow 6a+6\geq 3a^2 \Rightarrow \) \( \Rightarrow a^2-2a-2\leq 0 \Rightarrow (a-1)^2\leq 3 \Rightarrow a\leq \sqrt{3}+1<3 \). Contradictie!
Deci \( a^2>8 \) si \( \Delta<0 \). Deci ecuatia data nu are solutii.
Presupunem ca \( a^2\leq 8 \). Atunci avem
\( 6a+6=a^3\leq 8a \Rightarrow a\geq 3 \Rightarrow a^3\geq 3a^2\Rightarrow 6a+6\geq 3a^2 \Rightarrow \) \( \Rightarrow a^2-2a-2\leq 0 \Rightarrow (a-1)^2\leq 3 \Rightarrow a\leq \sqrt{3}+1<3 \). Contradictie!
Deci \( a^2>8 \) si \( \Delta<0 \). Deci ecuatia data nu are solutii.