IB Questionbank

User interface language: English | Español

Date	May 2022	Marks available	2	Reference code	22M.3.AHL.TZ2.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 2
Command term	Deduce	Question number	2	Adapted from	N/A

Question

This question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree $3$ and $4$ .

The cubic equation $x^{3} + p x^{2} + q x + r = 0$ , where $p, q, r \in ℝ$ , has roots $α, β$ and $γ$ .

Consider the equation $x^{3} - 7 x^{2} + q x + 1 = 0$ , where $q \in ℝ$ .

Noah believes that if $p^{2} \geq 3 q$ then $α, β$ and $γ$ are all real.

Now consider polynomial equations of degree $4$ .

The equation $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ , where $p, q, r, s \in ℝ$ , has roots $α, β, γ$ and $δ$ .

In a similar way to the cubic equation, it can be shown that:

$p = - (α + β + γ + δ)$

$q = α β + α γ + α δ + β γ + β δ + γ δ$

$r = - (α β γ + α β δ + α γ δ + β γ δ)$

$s = α β γ δ$ .

The equation $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = 0$ , has one integer root.

By expanding $(x - α) (x - β) (x - γ)$ show that:

$p = - (α + β + γ)$

$q = α β + β γ + γ α$

$r = - α β γ$ .

[3]

Show that $p^{2} - 2 q = α^{2} + β^{2} + γ^{2}$ .

[3]

b.i.

Hence show that ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} = 2 p^{2} - 6 q$ .

[3]

b.ii.

Given that $p^{2} < 3 q$ , deduce that $α, β$ and $γ$ cannot all be real.

[2]

Using the result from part (c), show that when $q = 17$ , this equation has at least one complex root.

[2]

By varying the value of $q$ in the equation $x^{3} - 7 x^{2} + q x + 1 = 0$ , determine the smallest positive integer value of $q$ required to show that Noah is incorrect.

[2]

e.i.

Explain why the equation will have at least one real root for all values of $q$ .

[1]

e.ii.

Find an expression for $α^{2} + β^{2} + γ^{2} + δ^{2}$ in terms of $p$ and $q$ .

[3]

f.i.

Hence state a condition in terms of $p$ and $q$ that would imply $x^{4} + p x^{3} + q x^{2} + r x + s = 0$ has at least one complex root.

[1]

f.ii.

Use your result from part (f)(ii) to show that the equation $x^{4} - 2 x^{3} + 3 x^{2} - 4 x + 5 = 0$ has at least one complex root.

[1]

State what the result in part (f)(ii) tells us when considering this equation $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12 = 0$ .

[1]

h.i.

Write down the integer root of this equation.

[1]

h.ii.

By writing $x^{4} - 9 x^{3} + 24 x^{2} + 22 x - 12$ as a product of one linear and one cubic factor, prove that the equation has at least one complex root.

[4]

h.iii.

Markscheme

attempt to expand $(x - α) (x - β) (x - γ)$ M1

$= (x^{2} - (α + β) x + α β) (x - γ)$ OR $= (x - α) (x^{2} - (β + γ) x + β γ)$ A1

$(x^{3} + p x^{2} + q x + r) = x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ$ A1

comparing coefficients:

$p = - (α + β + γ)$ AG

$q = (α β + β γ + γ α)$ AG

$r = - α β γ$ AG

Note: For candidates who do not include the AG lines award full marks.

[3 marks]

$p^{2} - 2 q = {(α + β + γ)}^{2} - 2 (α β + β γ + γ α)$ (A1)

attempt to expand ${(α + β + γ)}^{2}$ (M1)

$= α^{2} + β^{2} + γ^{2} + 2 (α β + β γ + γ α) - 2 (α β + β γ + γ α)$ or equivalent A1

$= α^{2} + β^{2} + γ^{2}$ AG

Note: Accept equivalent working from RHS to LHS.

[3 marks]

b.i.

EITHER

attempt to expand ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2}$ (M1)

$= (α^{2} + β^{2} - 2 α β) + (β^{2} + γ^{2} - 2 β γ) + (γ^{2} + α^{2} - 2 γ α)$ A1

$= 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α)$

$= 2 (p^{2} - 2 q) - 2 q$ or equivalent A1

$= 2 p^{2} - 6 q$ AG

attempt to write $2 p^{2} - 6 q$ in terms of $α, β, γ$ (M1)

$= 2 (p^{2} - 2 q) - 2 q$

$= 2 (α^{2} + β^{2} + γ^{2}) - 2 (α β + β γ + γ α)$ A1

$= (α^{2} + β^{2} - 2 α β) + (β^{2} + γ^{2} - 2 β γ) + (γ^{2} + α^{2} - 2 γ α)$ A1

$= {(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2}$ AG

Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.

[3 marks]

b.ii.

$p^{2} < 3 q \Rightarrow 2 p^{2} - 6 q < 0$

$\Rightarrow {(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} < 0$ A1

if all roots were real ${(α - β)}^{2} + {(β - γ)}^{2} + {(γ - α)}^{2} \geq 0$ R1

Note: Condone strict inequality in the R1 line.
Note: Do not award A0R1.

$\Rightarrow$ roots cannot all be real AG

[2 marks]

$p^{2} = {(- 7)}^{2} = 49$ and $3 q = 51$ A1

so $p^{2} < 3 q \Rightarrow$ the equation has at least one complex root R1

Note: Allow equivalent comparisons; e.g. checking $p^{2} < 6 q$

[2 marks]

use of GDC (eg graphs or tables) (M1)

$q = 12$ A1

[2 marks]

e.i.

complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).

a cubic curve always crosses the $x$ -axis at at least one point. R1

[1 mark]

e.ii.

attempt to expand ${(α + β + γ + δ)}^{2}$ (M1)

${(α + β + γ + δ)}^{2} = α^{2} + β^{2} + γ^{2} + δ^{2} + 2 (α β + α γ + α δ + β γ + β δ + γ δ)$ (A1)

$\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} = {(α + β + γ + δ)}^{2} - 2 (α β + α γ + α δ + β γ + β δ + γ δ)$

$(\Rightarrow α^{2} + β^{2} + γ^{2} + δ^{2} =) p^{2} - 2 q$ A1

[3 marks]

f.i.

$p^{2} < 2 q$ OR $p^{2} - 2 q < 0$ A1

Note: Allow FT on their result from part (f)(i).

[1 mark]

f.ii.

$4 < 6$ OR $2^{2} - 2 \times 3 < 0$ R1

hence there is at least one complex root. AG

Note: Allow FT from part (f)(ii) for the R mark provided numerical reasoning is seen.

[1 mark]

$(p^{2} > 2 q) (81 > 2 \times 24)$ (so) nothing can be deduced R1

Note: Do not allow FT for the R mark.

[1 mark]

h.i.

$- 1$ A1

[1 mark]

h.ii.

attempt to express as a product of a linear and cubic factor M1

$(x + 1) (x^{3} - 10 x^{2} + 34 x - 12)$ A1A1

Note: Award A1 for each factor. Award at most A1A0 if not written as a product.

since for the cubic, $p^{2} < 3 q (100 < 102)$ R1

there is at least one complex root AG

[4 marks]

h.iii.

Examiners report

The first part of this question proved to be very accessible, with the majority of candidates expanding their brackets as required, to find the coefficients $p, q$ and $r$ .

The first part of this question was usually answered well, though presentation in the second part sometimes left a lot to be desired. The expression $2 (p^{2} - 2 q) - 2 q$ was expected to be seen more often, as a 'pivot' to reaching the required result. Algebra was often lengthy, but untidily so, sometimes leaving examiners to do some mental tidying up on behalf of the candidate.

b.i.

[N/A]

b.ii.

A good number of candidates recognised the reasoning required in this part of the question and were able to score both marks.

Most candidates found applying this specific case to be very straightforward.

Most candidates offered incorrect answers in the first part; despite their working suggested utilisation of the GDC, it was clear that many did not appreciate what the question was asking. The second part was usually answered well, with the idea of complex roots occurring in conjugate pairs being put to good use.

e.i.

[N/A]

e.ii.

Some very dubious algebra was seen here, and often no algebra at all. Despite this, a good number of candidates seemed to make the 'leap' to the correct expression $p^{2} - 2 q$ , perhaps fortuitously so in a number of cases.

f.i.

[N/A]

f.ii.

Of those finding $p^{2} - 2 q$ in part f, a surprising number of answers seen employed the test of checking whether $p^{2} < 3 q$ .

Part i was usually not answered successfully, which may have been due to shortage of time. However, it was pleasing to see a number of candidates reach the end of the paper and successfully factorise the given quartic using a variety of methods. The final part required the $p^{2} < 3 q$ test. Though correct reasoning was sometimes seen, it was rare for this final mark to be gained.

h.i.

[N/A]

h.ii.

[N/A]

h.iii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Proof by induction, contradiction, counterexamples

Show 28 related questions

Hide related questions

Topic 1—Number and algebra

Question

Markscheme

Examiners report

Syllabus sections

View options