Conjugate roots and transformed equations · source reference

Question

$f(x)$ is a cubic polynomial with real coefficients. The diagram shows $y=f(x)$.

(a) What can be said about all roots of $f(x)=0$? [2]

Suppose $f(x)=2x^3-7x^2+16x+c$, with $c$ real.

(b) Show $c=-15$ if $f(x)=0$ has root $x=1-2i$. [1]

(c) Without using a calculator, determine the other roots. [3]

(d) Hence find roots of (i) $-15w^3+16w^2-7w+2=0$ [3], and (ii) $2iw^3-7w^2-16iw+15=0$ [2].

[2023/CJC/Prelim/I/4 modified]

Source working

Printed working

(a) There is only one positive real root of $f(x)=0$. Since all coefficients are real, the other two roots are a pair of complex conjugates.

(b) Substitute $x=1-2i$:

$$2(-11+2i)-7(-3-4i)+16(1-2i)+c=0,$$ $$15+c=0\Rightarrow c=-15.$$

(c) Conjugate root: $x=1+2i$.

$$\begin{aligned}2x^3-7x^2+16x-15&=[x-(1+2i)][x-(1-2i)](2x-k)\\&=[(x-1)+2i][(x-1)-2i](2x-k)\\&=(x^2-2x+5)(2x-k).\end{aligned}$$

Comparing constant terms, or by long division: $-5k=-15\Rightarrow k=3$. Last root $x=3/2$. Thus the three roots are $1+2i,1-2i,3/2$.

(d)(i) Replace $x$ with $1/w$:

$$2(1/w)^3-7(1/w)^2+16(1/w)-15=0,$$ $$2-7w+16w^2-15w^3=0.$$

Hence $1/w=1+2i,1-2i,3/2$, and

$$w=\frac1{1+2i}=\frac15-\frac25i,\quad\frac1{1-2i}=\frac15+\frac25i,\quad\frac23.$$

(d)(ii) Put $x=iw$:

$$2(iw)^3-7(iw)^2+16(iw)-15=0,$$ $$-2iw^3+7w^2+16iw-15=0\Rightarrow2iw^3-7w^2-16iw+15=0.$$

Thus $iw=1+2i,1-2i,3/2$, giving

$$w=\frac{1+2i}i=2-i,\quad\frac{1-2i}i=-2-i,\quad\frac3{2i}=-\frac32i.$$

Printed answer key

(a) One positive real root and two conjugate complex roots. (c) $1+2i,1-2i,3/2$. (d)(i) $1/5-2i/5,1/5+2i/5,2/3$. (d)(ii) $2-i,-2-i,-3i/2$.

Native diagrams

Provenance

2025 TJC H2 Math (9758) JC2 MSM: COMPLEX NUMBERS. Question pages 2; solution pages 8, 9. These are original document page numbers, retained as text metadata.