Polynomial problems (1 Viewer)

[byakuya kuchiki]

Suppose that b and d are real numbers and d /=/ (doesn't equal) 0. consider the polynomial P(z) = z^4 + bz^2 + d.

For what values of b does P(z) have a double root equal to root 3i.

anyone who has problems with polynomial questions are more than welcome to post here.

 

[Trebla]

Suppose that b and d are real numbers and d /=/ (doesn't equal) 0. consider the polynomial P(z) = z^4 + bz^2 + d.

For what values of b does P(z) have a double root equal to root 3i.

anyone who has problems with polynomial questions are more than welcome to post here.

If the polynomial has a double root of 3i then it must also have a double root of -3i since the polynomial is real.
Sum of roots in pairs:
(3i)(-3i) + (3i)(3i) + (3i)(-3i) + (-3i)(3i) + (-3i)(-3i) + (3i)(-3i) = 18
=> b = 18

 

[adomad]

do u dy/dz it and sub in 3i and then find b and then sub in b and z=3i to p(z)? i dunno haven't done 4U polynomials... just remember reading that in a 3U text book on repeated roots

 

[byakuya kuchiki]

but doesn't it say values? which means more than one value of b? i dunno. )':

 

[Trebla]

Another way to look at it, the factorised form of the quartic polynomial is:
(z - 3i)(z - 3i)(z + 3i)(z + 3i)
which simplifies to
(z - 3i)(z + 3i)(z - 3i)(z + 3i) = (z² + 9)²
= z⁴ + 18z² + 81

Hence b = 18

 

[gurmies]

Just an alternative solution.