Question

Fie ecuatia (m+3)x²-2(m-1)x+m-2=0
sa se determine m pt care ecuatia are solutii reale

Answer 1

[tex](m+3)x^2-2(m-1)x+m-2=0\\ Pentru\ a\ avea\ solutii\ reale\ \Delta\ trebuie\ sa\ fie\ \geq 0. Asadar:\\ {[2(m-1)]^2-4(m+3)(m-2) \geq 0}\\ 4(m^2-2m+1)-4(m^2+3m-2m-6)\geq 0\\ 4m^2-8m+4-4(m^2+m-6)\geq0\\ 4m^2-8m+4-4m^2-4m+24 \geq 0 \\ -12m+28\geq 0\\ 12m \leq 28 \\ m \leq \frac{7}{3}\\ S:m\in (-\infty,\frac{7}{3}][/tex]