Fie q -rația progresiei.
[tex]\it q=\dfrac{x+3}{3x-1}=\dfrac{9-x}{x-3}\ \Rightarrow (x+3)^2=(3x-1)(9-x) \Rightarrow\\ \\ \\ \Rightarrow x^2+6x+9=27x-3x^2-9+x \Rightarrow 4x^2+22x+18=0|_{:2} \Rightarrow \\ \\ \\ \Rightarrow 2x^2-11x+9=0 \Rightarrow 2x^2-2x-9x+9=0 \Rightarrow 2x(x-1)-9(x-1)=0 \Rightarrow \\ \\ \\ \Rightarrow (x-1)(2x-9)=0 \Rightarrow x_1=1,\ x_2=\dfrac{9}{2}[/tex]
