2010 H2 Mathematics Paper 2 Question 1

Complex Numbers

Answers

(i)

x=3+5i{x=3 + 5 \mathrm{i}} or x=3−5i{x=3 - 5 \mathrm{i}}

(ii)

a=−16,b=−20{a=- 16, b=- 20}
Other roots:−2−i,{- 2 - \mathrm{i}, } −2,{- 2, } 2{2}

Full solutions

(i)

x=−(−6)±(−6)2−4(1)(34)2(1)=6±−1002=6±10i2=3+5i   or   3−5i  ■\begin{align*} x &= \frac{-(- 6)\pm\sqrt{(- 6)^2-4(1)(34)}}{2(1)}\\ &= \frac{6\pm\sqrt{- 100}}{2}\\ \\ &= \frac{6\pm10\mathrm{i}}{2} \\ &= 3 + 5 \mathrm{i} \; \textrm{ or } \; 3 - 5 \mathrm{i} \; \blacksquare \end{align*}

(ii)

Since all coefficients are real, by the Conjugate Root Theorem, x=(−2+i)∗=−2−i{x= (- 2 + \mathrm{i})^* =- 2 - \mathrm{i}} is also a root
x4+4x3+x2+ax+b=(x−(−2+i))(x−(−2−i))(cx2+dx+e)=(x2+4x+5)(cx2+dx+e)\begin{align*} & x^4 + 4 x^3 + x^2 + ax + b \\ & = \Big(x-(- 2 + \mathrm{i})\Big)\Big(x-(- 2 - \mathrm{i})\Big)(cx^2+dx+e) \\ & = (x^2 + 4 x + 5)(cx^2+dx+e) \\ \end{align*}
Comparing coefficients,
x4:c=1x3:d+4c=4d=0x2:e+4d+5c=1e=−4x1:a=4e+5da=−16  ■x0:b=5eb=−20  ■\begin{align*} &x^4:&& c = 1 \\ &x^3:&& d+4c = 4 \\ && & d = 0 \\ &x^2:&& e+4d+5c = 1 \\ && & e = - 4 \\ &x^1:&& a = 4e+5d \\ && & a = - 16 \; \blacksquare \\ &x^0:&& b = 5e \\ && & b = - 20 \; \blacksquare \\ \end{align*}
cx2+dx+e=0x2−4=0(x+2)(x−2)=0x=−2   or x=2\begin{gather*} cx^2 + dx + e = 0 \\ x^2 - 4 = 0 \\ (x+2)(x-2) = 0 \\ x = -2 \; \textrm{ or } x = 2 \end{gather*}
Other roots:
x=−2−i,  x=−2,  x=2  ■x=- 2 - \mathrm{i}, \; x=- 2, \; x=2 \; \blacksquare