Question

Fara a rezolva ecuatia, sa se gaseasca suma patratelor radacinilor ecuatiei: (x²+2x)²-5(x²+2x)+3=0. Mersi

Answer 1

[tex](x^2+2x)^2-5(x^2+2x)+4=0\\ Notam\ x^2+2x=t\\ t^2-5t+4=0\\ t^2-4t-t+4=0\\ t(t-4)-(t-4)=0\\ (t-4)(t-1)=0 \Rightarrow t_1=4,t_2=1\\ Luam\ ambele\ cazuri\ posibile:\\ x^2+2x=4\\ x^2+2x-4=0\\ a=1,b=2,c=-4\\ x_1\cdot x_2=-4\\ x_1+x_2=-2 |()^2\\ (x_1+x_2)^2=4\\ x_{1}^2+x_{2}^2+2x_1\cdot x_2=4\\ x_{1}^2+x_{2}^2-8=4\\ x_{1}^2+x_{2}^2=\boxed{12}\\ Luam\ si\ al\ doilea\ caz:\\ x^2+2x=1\\ x^2+2x-1=0\\ a=1,b=2,c=-1\\ x_3\cdot x_4=-1\\ x_3+x_4=-2|()^2\\ (x_3+x_4)^2=4\\ [/tex]
[tex]x_{3}^2+x_{4}^2+2x_3\cdot x_4=4\\ x_{3}^2+x_{4}^2-2=4\\ x_{3}^2+x_{4}^2=\boxed{6}\\ x_{1}^2+x_{2}^2+x_{3}^2+x_{4}^2=6+12=\boxed{\boxed{18}}\\q.e.d[/tex]