How do you find all the zeros of P(x)=x^4+6x^2+9P(x)=x4+6x2+9?

1 Answer
Mar 4, 2016

x_1=x_3=i*sqrt(3), x_2=x_4=-i*sqrt(3)x1=x3=i3,x2=x4=i3

Explanation:

The polynomial is of a biquadratic kind so, making
x^2=yx2=y
And equating the polynomial to zero we have
y^2+6y+9=0y2+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)