2021 H2 Mathematics Paper 2 Question 1

Complex Numbers

Answers

a=−274{a=- \frac{27}{4}}
b=5{b=5}
Other roots: 1−12i,  −4{1 - \frac{1}{2} \mathrm{i}, \; - 4}

Full solutions

Since all the coefficients are real, by the conjugate root theorem, z=1−12i{z=1 - \frac{1}{2} \mathrm{i}} is also a root
x3+2x2+ax+b=(x−(1+12i))(x−(1−12i))(cx+d)=((x−1)2−(12i)2)(cx+d)=(x2−2x+54)(cx+d)\begin{align*} & x^3 + 2 x^2 + ax + b \\ & = \Big(x-\left({\textstyle 1 + \frac{1}{2} \mathrm{i}}\right)\Big)\Big(x-({\textstyle 1 - \frac{1}{2} \mathrm{i}})\Big)(cx+d) \\ & = \Big((x-1)^2 - \left({\textstyle \frac{1}{2}\mathrm{i}}\right)^2\Big)(cx+d) \\ & = \left(x^2 - 2 x + \frac{5}{4}\right)(cx+d) \\ \end{align*}
Comparing coefficients,
x3:c=1x2:d−2c=2d=4x:a=54c−2da=−274  ■x0:b=54db=5  ■\begin{align*} &x^3:& c &= 1 \\ &x^2:& d-2c &= 2 \\ && d &= 4 \\ &x:& a &= \frac{5}{4}c - 2d \\ && a &= - \frac{27}{4} \; \blacksquare \\ &x^0:& b &= \frac{5}{4}d \\ && b &= 5 \; \blacksquare \\ \end{align*}
x3+2x2−274x+5=0(x−(1+12i))(x−(1−12i))(x+4)=0\begin{gather*} x^3 + 2 x^2 - \frac{27}{4} x + 5 = 0 \\ \Big(x-\left({\textstyle 1 + \frac{1}{2} \mathrm{i}}\right)\Big)\Big(x-({\textstyle 1 - \frac{1}{2} \mathrm{i}})\Big)(x + 4) = 0 \\ \end{gather*}
Other roots:
x=1−12i,x=−4  ■x={\textstyle 1 - \frac{1}{2} \mathrm{i}}, \quad x=- 4 \; \blacksquare