Question

биквадрат y4-3y2+2=0

Answer 1

Пусть у²=t
t²-3t+2=0
D=9-4*1*2=1
t1=(3+1)/2=2
t2=(3-1)/2=1
y²=2
y1=√2, y2=-√2
y²=1
y3=1, y4=-1