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