Пусть x^2 + 2x = t, тогда
t(t + 2) = 3
t^2 + 2t - 3 = 0
D = 4 + 12 = 16 = 4^2
t₁= ( - 2 + 4)/2 = 1;
t₂ = ( - 2 - 4)/2 = - 3;
x^2 + 2x = 1
x^2 + 2x - 1 = 0
D = 4 + 4 = 8
x₁= ( - 2 + 2√2)/2 = - 1 + √2;
x₂ = ( - 2 - 2√2)/2 = - 1 - √2;
x^2 + 2x = - 3
x^2 + 2x + 3 = 0
D = 4 - 4*3 < 0
∅ нет реш
Ответ
- 1 - √2;
- 1 + √2