Answer:
Ziros is : x=+-sqrt(6) and x=+-3i
Step-by-step explanation:
x^4+3x^2-54=0
t=x^2, then
t^2+3t-54=0
a=1, b=3, c=-54 (Quadratic equation coefficients)
t1 = (-b+sqrt(b^2-4ac) /2a
t1 = (-3+sqrt(3^2 - 4*1*(-54))/2*1
t1 = (-3+sqrt(9+216))/2
t1 = (-3+sqrt(225))/2
t1 = (-3+15)/2
t1 = 12/2
t1 =6
t2 = (-b-sqrt(b^2-4ac)/2a
t2 =(-3-15)/2
t2=(-18)/2
t2=-9
t1=x^2 =6
x=+-sqrt(6)
t2=x^2=-9
x=+-sqrt(-9)... (-1=i^2)
x=+-sqrt(9i^2)
x=+-3i
