Question

The cubiz equation has real coefficients. If two of the roots are and
(i) State the third root
(ii) Find the cubic equation in the form

Answer 1

P(z) = 0 therefore we can find the factors from the roots

factors of root -3 == (x + 3)

factors of root -1 -3i = (x + 1 + 3i)

Question I : State the third root of the equation

the third root would be the conjugate of the complex root which is (-1 + 3i)

therefore we get our third factor which is (x + 1 - 3i)

Question II : Find the cubic equation in the form

with that equation the coefficients must be real, we can multiply our factors

(x + 3)(x + 1 + 3i)(x + 1 - 3i) = 0

= 0

= 0

Maaf kalau salah, sudah lama tidak mengerjakan soal seperti ini.

Answer 2

I. Roots of -3 : the factor is (x+3), roots of -1-3i : factor is (x+(1+3i))

One of the roots is real, and other one is imaginary, if one root is imaginary that means another one is also imaginary according to quadratic formula

quadratic formula :

x = (-b ± √(D))/2a , D = b²-4ac

one of the imaginary root is -1-3i (or -(1+3i)), meaning the other one must be it's conjugate that is -1+3i (or -(1-3i)) (factor is (x+(1-3i))

ii. The cubic equation in form of roots are :

0 = (x+3)(x+(1-3i))(x+(1+3i))

0 = (x+3)(x²+2x+10)

0 = x³+5x²+16x+30