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Date	May 2015	Marks available	5	Reference code	15M.1.hl.TZ2.12
Level	HL only	Paper	1	Time zone	TZ2
Command term	Determine, Hence, and Show that	Question number	12	Adapted from	N/A

Question

The cubic equation ${x^3} + p{x^2} + qx + c = 0$, has roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $. By expanding $(x - \alpha )(x - \beta )(x - \gamma )$ show that

(i) $p = - (\alpha + \beta + \gamma )$;

(ii) $q = \alpha \beta + \beta \gamma + \gamma \alpha $;

(iii) $c = - \alpha \beta \gamma $.

[3]

It is now given that $p = - 6$ and $q = 18$ for parts (b) and (c) below.

(i) In the case that the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $ form an arithmetic sequence, show that one of the roots is $2$.

(ii) Hence determine the value of $c$.

[5]

In another case the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $ form a geometric sequence. Determine the value of $c$.

[6]

Markscheme

(i)-(iii) given the three roots $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $, we have

${x^3} + p{x^2} + qx + c = (x - \alpha )(x - \beta )(x - \gamma )$ M1

$ = \left( {{x^2} - (\alpha + \beta )x + \alpha \beta } \right)(x - \gamma )$ A1

$ = {x^3} - (\alpha + \beta + \gamma ){x^2} + (\alpha \beta + \beta \gamma + \gamma \alpha )x - \alpha \beta \gamma $ A1

comparing coefficients:

$p = - (\alpha + \beta + \gamma )$ AG

$q = (\alpha \beta + \beta \gamma + \gamma \alpha )$ AG

$c = - \alpha \beta \gamma $ AG

[3 marks]

METHOD 1

(i) Given $ - \alpha - \beta - \gamma = - 6$

And $\alpha \beta + \beta \gamma + \gamma \alpha = 18$

Let the three roots be $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $

So $\beta - \alpha = \gamma - \beta $ M1

or $2\beta = \alpha + \gamma $

Attempt to solve simultaneous equations: M1

$\beta + 2\beta = 6$ A1

$\beta = 2$ AG

(ii) $\alpha + \gamma = 4$

$2\alpha + 2\gamma + \alpha \gamma = 18$

$ \Rightarrow {\gamma ^2} - 4\gamma + 10 = 0$

$ \Rightarrow \gamma = \frac{{4 \pm {\text{i}}\sqrt {24} }}{2}$ (A1)

Therefore $c = - \alpha \beta \gamma = - \left( {\frac{{4 + {\text{i}}\sqrt {24} }}{2}} \right)\left( {\frac{{4 - {\text{i}}\sqrt {24} }}{2}} \right)2 = - 20$ A1

METHOD 2

(i) let the three roots be $\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d$ M1

adding roots M1

to give $3\alpha = 6$ A1

$\alpha = 2$ AG

(ii) $\alpha $ is a root, so ${2^3} - 6 \times {2^2} + 18 \times 2 + c = 0$ M1

$8 - 24 + 36 + c = 0$

$c = - 20$ A1

METHOD 3

(i) let the three roots be $\alpha ,{\text{ }}\alpha - d,{\text{ }}\alpha + d$ M1

adding roots M1

to give $3\alpha = 6$ A1

$\alpha = 2$ AG

(ii) $q = 18 = 2(2 - d) + (2 - d)(2 + d) + 2(2 + d)$ M1

${d^2} = - 6 \Rightarrow d = \sqrt 6 {\text{i}}$

$ \Rightarrow c = - 20$ A1

[5 marks]

METHOD 1

Given $ - \alpha - \beta - \gamma = - 6$

And $\alpha \beta + \beta \gamma + \gamma \alpha = 18$

Let the three roots be $\alpha ,{\text{ }}\beta ,{\text{ }}\gamma $.

So $\frac{\beta }{\alpha } = \frac{\gamma }{\beta }$ M1

or ${\beta ^2} = \alpha \gamma $

Attempt to solve simultaneous equations: M1

$\alpha \beta + \gamma \beta + {\beta ^2} = 18$

$\beta (\alpha + \beta + \gamma ) = 18$

$6\beta = 18$

$\beta = 3$ A1

$\alpha + \gamma = 3,{\text{ }}\alpha = \frac{9}{\gamma }$

$ \Rightarrow {\gamma ^2} - 3\gamma + 9 = 0$

$ \Rightarrow \gamma = \frac{{3 \pm {\text{i}}\sqrt {27} }}{2}$ (A1)(A1)

Therefore $c = - \alpha \beta \gamma = - \left( {\frac{{3 + {\text{i}}\sqrt {27} }}{2}} \right)\left( {\frac{{3 - {\text{i}}\sqrt {27} }}{2}} \right)3 = - 27$ A1

METHOD 2

let the three roots be $a,{\text{ }}ar,{\text{ }}a{r^2}$ M1

attempt at substitution of $a,{\text{ }}ar,{\text{ }}a{r^2}$ and $p$ and $q$ into equations from (a) M1

$6 = a + ar + a{r^2}\left( { = a(1 + r + {r^2})} \right)$ A1

$18 = {a^2}r + {a^2}{r^3} + {a^2}{r^2}\left( { = {a^2}r(1 + r + {r^2})} \right)$ A1

therefore $3 = ar$ A1

therefore $c = - {a^3}{r^3} = - {3^3} = - 27$ A1

[6 marks]

Total [14 marks]