Question

determinati valorile parametrului real m astfel incat acuatia xpatrat -4mx+1=0 sa aiba solutii reale

Answer 1

[tex]x^2-4mx+1=0\\ \text{O ecuatie de gradul 2 are solutii reale atunci cand delta e mai mare }\\ \text{ sau egal cu 0.}\\ \Delta=16m^2-4\\ \Delta\geq 0\\ 16m^2-4\geq 0\\ 16m^2\geq 4\\ m^2\geq \dfrac{1}{4}\\ |m|\geq \dfrac{1}{2}\\ S:m\in \left(-\inf; -\dfrac{1}{2}\right]\cup \left[\dfrac{1}{2}; \inf\right)[/tex]