[tex]Pentru \ ca \ ecuatia \ ta \ sa \ aibe \ 2 \ solutii \ distincte, \ \Delta \ trebuie \ sa \ fie \ mai \ mare \ decat \ 0\\ \\x^2-(m+3)x+m^2=0\\ \\ \Delta=[-(m+3)]^2-4\cdot 1 \cdot m^2=m^2+6m+9-4m^2=-3m^2+6m+9\\ \\Avem \ functia \ de\ gradul \ 2: \ f(m)=-3m^2+6m+9, \\ \\ Observam \ ca \ a_m=-3<0 \ si \ \Delta=6^2-4\cdot 9\cdot(-3)=36+108=144>0\\ \\ \Rightarrow \ f(m)>0 \ cand \ m \in (m_1, \ m_2) \\ \\ m_{1, \ 2}=\frac{-6\pm \sqrt{144}}{2\cdot(-3)}=\frac{-6 \pm 12}{-6}\\ \\ m_2=\frac{-6-12}{-6}=\frac{-18}{-6}=3\\ \\ m_1=\frac{-6+12}{-6}=\frac{6}{-6}=-1\\ \\ f(m)>0 \ cand \ m \in (-1; \ 3)\\ \\ \\ \\ \boxed{Raspuns: \ e)}[/tex]
