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

the opening of this porabola is facing up

step-by-step explanation:

step-by-step explanation: