Question

Решить уравнение x^4-39x^2+108=0

Answer 1

$x^2=y \ y^2-39y+108=0 \ D=b^2-4ac=1521-432=1089 \ y1=(39-33)/2=3 \ y2=(39+33)/2=36 \ x1=+- sqrt{3} \ x2 = + - 6$
Ответ:
$x = - sqrt{3}, sqrt{3}, -6 , 6.$

Answer 2

x^4-39x^2+108=0

x^2=t

t^2-39t+108

За Виета t1=36 t2=3

x^2=36

x^2=6^2

x1=6

x^2=3

x2=√3