How do you find all the zeros of #P(x)=x^4+6x^2+9#?

1 Answer
Mar 4, 2016

Answer:

#x_1=x_3=i*sqrt(3), x_2=x_4=-i*sqrt(3)#

Explanation:

The polynomial is of a biquadratic kind so, making
#x^2=y#
And equating the polynomial to zero we have
#y^2+6y+9=0#

#Delta=-36-36=0#
#y=(-6+-0)/2# => #y_1=y_2=-3#
Then
#x_(1,2)=sqrt(y_1)=sqrt(-3)# => #x_1=i*sqrt(3)#, #x_2=-i*sqrt(3)#
#x_(3,4)=sqrt(y_2)=sqrt(-3)# => #x_3=i*sqrt(3)#, #x_4=-i*sqrt(3)#
So
#x_1=x_3=i*sqrt(3)#
#x_2=x_4=-i*sqrt(3)#