Question

sa se determine mapartine lui R astfel incat solutiile x1si x2 ale ecuatiei xpatrat-2x+6m-1=0sa verifice relatia x1-x2=x1x2​

Answer 1

$x^2-2x+6m-1 = 0\\ \\ \Delta \geq 0 \Rightarrow 4-4(6m-1) \geq 0\Rightarrow 6m-1\leq 1\Rightarrow \\ \\ \Rightarrow m \leq \frac{1}{3} \\\\\text{Abscisa varfului parabolei este:}\\-\frac{b}{2a} = -\frac{-2}{2} = 1 \Rightarrow \frac{|x_2-x_1|}{2} = 1 \Rightarrow |x_1-x_2| = 2 \Rightarrow\\ \\ \Rightarrow x_1-x_2 = \pm 2 \Rightarrow \pm 2 = x_1x_2 \Rightarrow \pm 2 = 6m-1 \Rightarrow\\ \\ \Rightarrow 6m = \pm 2+1 \Rightarrow 6m \in \{-1; 3\} \Rightarrow m\in \left\{-\frac{1}{6}; \frac{1}{2}\right\}$

$\text{dar }m\leq \frac{1}{3} \Rightarrow \boxed{m = -\frac{1}{6}}$