POlynomials q (1 Viewer)

[undalay]

for the polynomial p(z) = z^4 + bz^2 + (b^2/4)

for what values of b does p(z) have real roots?

 

[cwag]

b must be less than or equal to 0 for p(z) to have real roots.

consider it graphically, if b is a postive number, there is no posible way that P(z) will cross the x axis. when b is 0 x^4 has one real root. when b <0 P(z) has two real roots.

 

[samwell]

make x=z^2 and then factorise:
x^2 + bx + b^2/4
4x^2 + 4bx + b^2
then make b^2- 4ac>/= to 0
B=4b and a=4 and c=b^2
therefore delta =/ > 0 and b is an element of all real values.

 

[Iruka]

z^4 + bz^2 + (b^2/4) is a perfect square.

So factorise it.

z^4 + bz^2 + (b^2/4) = (z^2 + b/2)^2

Thus if p(z) = 0, z^2 + b/2 = 0, so z= +/- sqrt(-b/2), so if z is to be real, then, as cwag said, b<= 0.

 

[undalay]

thx guys.