Question

Решите уравнение (x^2+x)^2-8(x^2+x)+12=0 ^ степень

Answer 1

$( x^{2} +x) ^{2} -8( x^{2} +x)+12=0 \ x^{2} +x=y \ y ^{2} -8y+12=0 \ D>0 \ y_{1} =6 \ y _{2} =2 \ x^{2} +x=6 \ x^{2} +x-6=0 \ D>0 \ x_{1} =2 \ x_{2} =-3 \ x^{2} +x=2 \ x^{2} +x-2=0 \ D>0 \ x_{3} =1 \ x_{4} =-2$