Find all the zeros

x^4+3x^2-54=0

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