[tex]x^2-mx-3 = 0 \quad $si$\quad x^2-(m+7)x+4 = 0 \\ $(au o radacina comuna)$ \\ \\ $Facem sistem, si aflam solutiile comune si punem dupa aceea conditiile\\ pentru m.
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[tex]\left\{ \begin{array}{c} x^2-mx-3=0\quad \quad \quad \quad \quad $ $ $ $ \\ x^2-(m+7)x+4= 0 \Big|\cdot (-1)\end{array} \right \Rightarrow \left\{ \begin{array}{c} x^2-mx-3=0 \quad \quad \\ -x^2+(m+7)x-4= 0 \end{array} \right \Rightarrow\\ ~ \quad \quad \quad \quad \quad \quad \quad \quad\quad \quad \quad \quad\quad \quad \quad \quad\quad \quad -----------(+)\\ \\ \Rightarrow 0-mx+(m+7)x-7 = 0 \Rightarrow x(-m+m+7) = 7 \Rightarrow [/tex]
[tex] \Rightarrow x\cdot 7 = 7 \Rightarrow x = \dfrac{7}{7} \Rightarrow \boxed{x = 1}\rightarrow \text{Solutie unica, independenta de m.}\\ \\ \Rightarrow \boxed{m \in \mathbb{R}}\rightarrow solutie ~ finala.[/tex]
