Nine v.S.
Revision 20
Exercise 2
a) Show that the complex number 𝑖 is a root of the equation
𝑥 ! − 5𝑥 ! + 7𝑥 ! − 5𝑥 + 6 = 0
Plug in 𝑖 for x:
→ 𝑖 ! − 5𝑖 ! + 7𝑖 ! − 5𝑖 + 6
= 1 − 5(−𝑖) + 7(−1) − 5𝑖 + 6
= 1 + 5𝑖 − 7 − 5𝑖 + 6
=0
b) Find the other roots of this equation.
The conjugate of 𝑖 is – 𝑖 → it is a root
𝑥 − 𝑖 𝑥 + 𝑖 = 𝑥! + 1
Long division:
(𝑥 ! − 5𝑥 ! + 7𝑥 ! − 5𝑥 + 6) ÷ (𝑥 ! + 1)
x 2 − 5x + 6
x − 5x + 7x − 5x + 6
x 2 +1
4
3
2
𝑥! + 𝑥!
−5𝑥 ! + 6𝑥 !
−5𝑥 ! − 5𝑥
6𝑥 ! + 6
6𝑥 ! + 6
0
!
→ 𝑥 − 5𝑥 + 6 = (𝑥 − 2)(𝑥 + 3)
→𝑥 = 2 𝑜𝑟 𝑥 = 3
Hence, the roots of 𝑥 ! − 5𝑥 ! + 7𝑥 ! − 5𝑥 + 6 = 0 are:
𝑥 = 𝑖 𝑥 = −𝑖
𝑥 = 2 𝑥=3