[tex]\displaystyle Daca~notam~t= \sqrt[3]{x^2+8},~ecuatia~devine~ \\ \\ mt^2+2(m-1)t+m-1=0.~(*) \\ \\ Sa~observam~ca~t~ia~orice~valoare \ge 2.~Deci~o~conditie~necesara~si \\ \\ si~suficienta~pentru~ca~toate~radacinile~sa~fie~reale~ar~fi \\ \\ t_1 \ge 2 ~si~ t_2 \ge 2. \\ \\ Sa~mai~observam~ca~m\ \textgreater \ 0.~(In~caz~contrar~am~avea~mt^2\ \textless \ 0~; \\ \\ 2(m-1)t\ \textless \ 0~si~m-1\ \textless \ 0)~deci~egalitatea~(*)~ar~fi~imposibila). \\ \\ Cea~mai~rapida~cale~prin~care~am~demonstrat~ca~m \le \frac{5}{9}~a~fost \\ \\ urmatoarea: [/tex]
[tex]\displaystyle (*) \Leftrightarrow m(t^2+2t+1)-2t-1=0 \Leftrightarrow \\ \\ \Leftrightarrow m(t+1)^2=2t+1 \Leftrightarrow \\ \\ \Leftrightarrow m= \frac{2t+1}{(t+1)^2}.~(**) \\ \\ Consideram~functia~f: [2, + \infty) \to \mathbb{R},~f(x)= \frac{2x+1}{(x+1)^2}. \\ \\ Avem~f'(x)= -\frac{2x}{(x+1)^3}\ \textless \ 0,~deci~f~este~strict~descrescatoare. \\ \\ Din~(**)~avem~m=f(t) \le f(2)= \frac{5}{9}.[/tex]
[tex]\displaystyle Metoda~prin~care~am~ajuns~la~contradictie: \\ \\ mt^2+2(m-1)t+m-1=0~(*) \\ \\ \Delta=4(1-m) \\ \\ Observam~ca~nu~putem~avea~t_1=t_2.~(ar~implica~ \Delta =0,~deci~m=1, \\ \\ deci~t=0~(nu~convine) \\ \\ [/tex]
[tex]\displaystyle Pentru~a~avea~t_1,~t_2 \ge 2~am~pus~conditia~(necesara,~dar~nu~si \\ \\ suficienta!)~(t_1-2)(t_2-2) \ge 0.~Nu~este~o~conditie~suficienta, \\ \\ deoarece~este~verificata~de~ \left \{ {{t_1 \ge 2} \atop {t_2 \ge 2}} \right.~si~ \left \{ {{t_1 \le 2} \atop {t_2 \le 2}} \right. . \\ \\ Pentru~a~elimina~al~doilea~sistem,~completam~conditia~cu~ \\ \\ t_1+t_2\ \textgreater \ 4.~(egalitatea~nu~poate~avea~loc,~conform~afirmatiilor \\ \\ anterioare~;~oricum...~se~pare~ca~aceasta~conditie~nu~este~utila). \\ \\ [/tex]
[tex]\displaystyle (t_1-2)(t_2-2) \ge 0 \Leftrightarrow t_1t_2-2(t_1+t_2)+4 \ge 0.~(***) \\ \\ t_1t_2= \frac{m-1}{m} \\ \\ t_1+t_2= \frac{2(1-m)}{m} \\ \\ Deci~(***) \Leftrightarrow 5 \cdot \frac{m-1}{m}+4 \ge 0,~deci~9- \frac{5}{m} \ge 0 ,~deci~m \ge \frac{5}{9}. \\ \\ Din~m \le \frac{5}{9} ~si~ m \ge \frac{5}{9}~rezulta~m= \frac{5}{9}. \\ \\ Testand~acet~caz,~constatam~ca~nu~convine~(o~radacina \\ \\ va~fi~negativa). \\ \\ Prin~urmare~raspunsul~problemei~este~e).[/tex]
