[tex]\displaystyle a)~Conditia~de~existenta~a~radicalului:~x+6 \ge 0 \Leftrightarrow x \ge -6. \\ \\ Totodata~x= \sqrt{x+6} \ge 0. ~Din~ultimele~doua~relatii~rezulta \\ \\ x \ge 0.\\ \\ Prin~ridicare~la~patrat,~obtinem:~x+6=x^2 \Leftrightarrow x^2-x-6=0. \\ \\ \Delta=25 \\ \\ x_1=\frac{1-\sqrt{25}}{2}=\frac{1-5}{2}=-2,~nu~convine!~(caci~x \ge 0) \\ \\ x_2=\frac{1+\sqrt{25}}{2}=\frac{1+5}{2}=3. \\ \\ Deci~solutia~este~3.[/tex]
[tex]\displaystyle b)~Notam ~3^x=t\ \textgreater \ 0. \\ \\ Ecuatia~devine:~3^x+(3^x)^2=12 \Leftrightarrow t+t^2=12 \Leftrightarrow t^2+t-12=0. \\ \\ \Delta=49. \\ \\ t_1= \frac{-1- \sqrt{49}}{2}=-4,~nu~convine!~(caci~t\ \textgreater \ 0) \\ \\ t_2= \frac{-1+ \sqrt{49}}{2}=3. \\ \\ Deci~t=3 \Rightarrow 3^x=3 \Rightarrow \boxed{x=1}~. [/tex]
[tex]Observatie:~Functia~f: \mathbb{R} \rightarrow \mathbb{R},~f(x)=3^x+9^x~este~strict~ \\ \\ crescatoare.~Prin~urmare,~ecuatia~f(x)=12~are~cel~mult~o \\ \\ solutie~reala.~Cum~f(1)=12,~rezulta~ca~1~este~unica~solutie~a \\ \\ ecuatiei.[/tex]
