Răspuns :
Ecuatia se poate scrie: [tex](x^2-x)(x^2-x-2)=24[/tex]
Facem notatia: [tex]t=x^2-x[/tex]
Ecuatia devine: [tex]t(t-2)=24[/tex]
Este o ecuatie de grad 2, care se rezolva pentru a obtine 2 solutii:
[tex]t^2-2t-24=0\\ t^2-6t+4t-24=0\\(t-6)(t+4)=0\\ \Rightarrow t_1=6, t_2=-4.[/tex]
Luam fiecare caz in parte:
Cazul 1: [tex]t=6[/tex] .
[tex]x^2-x=6\\ \\ x^2-x-6=0\\ x^2-3x+2x-6=0\\ (x-3)(x+2)=0\\ \\ \Rightarrow x_1=3, \ \ x_2=-2.[/tex]
Cazul 2: [tex]t=-4[/tex] .
[tex]x^2-x=-4\\ \\ x^2-x+4=0\\ \Delta = -15\\ \\ \Rightarrow x_3=\dfrac{1+i\sqrt{15}}{2}, \ \ x_4=\dfrac{1-i\sqrt{15}}{2}.[/tex]
In concluzie, multimea solutiilor este:
[tex]S=\left\{-2,3,\dfrac{1-i\sqrt{15}}{2}, \dfrac{1+\sqrt{15}}{2}\right\}.[/tex]
Facem notatia: [tex]t=x^2-x[/tex]
Ecuatia devine: [tex]t(t-2)=24[/tex]
Este o ecuatie de grad 2, care se rezolva pentru a obtine 2 solutii:
[tex]t^2-2t-24=0\\ t^2-6t+4t-24=0\\(t-6)(t+4)=0\\ \Rightarrow t_1=6, t_2=-4.[/tex]
Luam fiecare caz in parte:
Cazul 1: [tex]t=6[/tex] .
[tex]x^2-x=6\\ \\ x^2-x-6=0\\ x^2-3x+2x-6=0\\ (x-3)(x+2)=0\\ \\ \Rightarrow x_1=3, \ \ x_2=-2.[/tex]
Cazul 2: [tex]t=-4[/tex] .
[tex]x^2-x=-4\\ \\ x^2-x+4=0\\ \Delta = -15\\ \\ \Rightarrow x_3=\dfrac{1+i\sqrt{15}}{2}, \ \ x_4=\dfrac{1-i\sqrt{15}}{2}.[/tex]
In concluzie, multimea solutiilor este:
[tex]S=\left\{-2,3,\dfrac{1-i\sqrt{15}}{2}, \dfrac{1+\sqrt{15}}{2}\right\}.[/tex]