Question

Fie ecuatia $m^{2} x+1=m(x+1)$ . Sa se discute,in functe de parametrul real m,natura solutiilor ecuatiei. Pentru ce valori ale lui m ecuatia are solutia $\frac{1}{2}$ ?

Answer 1

m²x+1=mx+m, m²x-mx=m-1; xm(m-1)=m-1; x=(m-1)/m(m-1)⇒ x=1/m. m≠0. m∈ R-{0}; 1/2=1/m ⇒m=2

Answer 2

[tex]m^2x+1=m(x+1) \Rightarrow m^2x+1=mx+m \Rightarrow m^2x+1 -mx-m=0 \\\;\\ \Rightarrow mx(m-1)-(m-1)=0 \Rightarrow mx(m-1) = (m-1) \\\;\\\Rightarrow x=\dfrac{m-1}{m(m-1)} \Rightarrow x = \dfrac{1}{m} \ solutie \ unica\ \ \forall m\in \mathbb{R}-\{0, 1\}. \\\;\\ Daca\ m = 0,\ ecuatia\ devine: \\\;\\ 0x+1=0 \Rightarrow 0x = -1 \Rightarrow\ ecuatie\ imposibila \Rightarrow S = \phi \\\;\\ Daca\ \ m =1, \ ecuatia\ devine: \\\;\\ x +1 =x+1 \Rightarrow\ x-x = 1-1 \Rightarrow\ 0x=0,\ \ ecuatie\  nedeterminata\Rightarrow S=\mathbb{R}[/tex]

Daca ecuatia admite solutia x = 1/2, vom avea:

[tex]m^2\cdot\dfrac{1}{2}+1 =m\left(\dfrac{1}{2} + 1\right) \Rightarrow m^2\cdot\dfrac{1}{2} +1=m\cdot\dfrac{3}{2} \Rightarrow m^2+2=3m \Rightarrow \\\;\\ \Rightarrow m^2-3m+2=0 \Rightarrow m_1=1,\ \ m_2=2[/tex]