Conjugate roots and transformed equations

Given cubic graphSupplied qualitative graph of y=f(x), with labelled x and y axes and origin O.xyO
Question diagram 1
Given cubic graphSupplied qualitative graph of y=f(x), with labelled x and y axes and origin O.xyO

Supplied qualitative cubic graph of y=f(x), with the x-axis, y-axis and origin O.

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

(a) What can be said about all roots of f(x)=0f(x)=0? marks

Suppose f(x)=2x3−7x2+16x+cf(x)=2x^3-7x^2+16x+c, with cc real.

(b) Show c=−15c=-15 if f(x)=0f(x)=0 has root x=1−2ix=1-2i. mark

(c) Without using a calculator, determine the other roots. marks

(d) Hence find roots of (i) −15w3+16w2−7w+2=0-15w^3+16w^2-7w+2=0 marks, and (ii) 2iw3−7w2−16iw+15=02iw^3-7w^2-16iw+15=0 marks.

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

Total marks: 11

Solution

(a)

Describe the roots of a real cubic

f(x)f(x) is a cubic polynomial with real coefficients, and the graph of y=f(x)y=f(x) crosses the xx-axis once, at a positive xx. What can be said about all roots of f(x)=0f(x)=0?

Hint

  1. Count the real roots from the graph.

  2. Non-real roots of real polynomials come in conjugate pairs.

Answer · (a)

One positive real root and a nonreal conjugate pair.

(b)

Find a missing coefficient

f(x)=2x3−7x2+16x+cf(x)=2x^3-7x^2+16x+c with cc real. Show that c=−15c=-15 if x=1−2ix=1-2i is a root.

Explore this idea

Hint

  1. Substitute x=1−2ix=1-2i and set the result to 00.

Answer · (b)

c=−15c=-15.

Explore the idea

The roots as c changes

The cubic 2x3−7x2+16x+c2x^3-7x^2+16x+c is always increasing, so it keeps exactly one real root for every cc. The other two stay a conjugate pair.

Value of c-15
Try an example
root 1
1 + 2i
root 2
1 − 2i
root 3
1.5
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Complex roots of real polynomials come as mirror images in the real axis, so a cubic can never have all three non-real roots.

Ideas to try
  1. Set c=−15c=-15: the roots are 1±2i1\pm2i and 1.51.5.
  2. Slide cc across its whole range. One root stays on the real axis, and the other two stay mirror images of each other.
  3. Try the Spacetime view to see each root trace a path as cc increases.
(c)

Factor using a conjugate pair

Without a calculator, determine the other roots of f(x)=0f(x)=0.

Hint

  1. Write down the conjugate root.

  2. Build the quadratic factor and divide.

Answer · (c)

Other roots 1+2i,3/21+2i,3/2.

(d)(i)

Reciprocal roots

Hence find the roots of −15w3+16w2−7w+2=0-15w^3+16w^2-7w+2=0.

Explore this idea

Hint

  1. Compare with w3f(1/w)w^3f(1/w).

Answer · (d)(i)

(1+2i)/5,(1−2i)/5,2/3(1+2i)/5,(1-2i)/5,2/3.

Explore the idea

Taking reciprocals

Slide from the roots xx of ff to their reciprocals. Moduli become 1∣x∣\frac1{|x|} and arguments change sign.

From x to 1/x0
Try an example
point 1
1 − 2i
point 2
1 + 2i
point 3
1.5
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Reciprocals reflect across the unit circle and in the real axis: a large root becomes a small one.

Ideas to try
  1. At the end of the slide the points are 1∓2i5≈0.2∓0.4i\frac{1\mp2i}5\approx0.2\mp0.4i and 23\frac23. All of them moved across the unit circle.
  2. Points outside the unit circle land inside, and conversely.
(d)(ii)

Rotated roots

Hence find the roots of 2iw3−7w2−16iw+15=02iw^3-7w^2-16iw+15=0.

Explore this idea

Hint

  1. Compare with −f(iw)-f(iw).

Answer · (d)(ii)

2−i,−2−i,−3i/22-i,-2-i,-3i/2.

Explore the idea

Turning the roots through −90°

Multiplying every root of ff by −i-i rotates the whole picture a quarter-turn clockwise.

Rotation0°
Try an example
point 1
1 − 2i
point 2
1 + 2i
point 3
1.5
  • Moves as the slider changes; a faint path shows where it goes
Key idea

Substituting iwiw for xx rotates every root by −90∘-90^\circ, because w=−ixw=-ix.

Ideas to try
  1. Slide to −90∘-90^\circ. The points land on 2−i2-i, −2−i-2-i and −32i-\frac32i.
  2. The shape and the sizes never change: a rotation preserves distances.
Takeaway

Transform the roots, not the work

Equations built from f(1/w)f(1/w) or f(iw)f(iw) have roots obtained from the roots of ff by simple maps. Identify the map, then apply it.