2018 H2 Maths Paper 2 Q2 — Worked Solution

Part (a)

5 marks

Since $s$ and $t$ are real, the coefficients of the quartic are real, so complex roots occur in conjugate pairs. As $2 - 3i$ is a root, $2 + 3i$ is also a root.

\left( x - (2 - 3i) \right)\left( x - (2 + 3i) \right)

= \left( (x - 2) + 3i \right)\left( (x - 2) - 3i \right)

= (x - 2)^{2} + 9

= x^{2} - 4x + 13

Approach

The quartic has leading coefficient $4$ , and $x^{2} - 4x + 13$ (the factor from the known conjugate pair) is monic, so the remaining quadratic factor must also be monic. We write the quartic as $4\left( x^{2} - 4x + 13 \right)\left( x^{2} + bx + c \right)$ and compare coefficients to find $b$ and $c$ , then $s$ and $t$ .

4x^{4} - 20x^{3} + sx^{2} - 56x + t = 4\left( x^{2} - 4x + 13 \right)\left( x^{2} + bx + c \right)

Expanding the right side,

4\left( x^{2} - 4x + 13 \right)\left( x^{2} + bx + c \right)

= 4x^{4} + 4bx^{3} + 4cx^{2} - 16x^{3} - 16bx^{2} - 16cx + 52x^{2} + 52bx + 52c

= 4x^{4} + (4b - 16)x^{3} + (4c - 16b + 52)x^{2} + (52b - 16c)x + 52c

Comparing the coefficient of $x^{3}$ with $- 20$ ,

4b - 16 = - 20

\Rightarrow b = - 1

Comparing the coefficient of $x$ with $- 56$ ,

52b - 16c = - 56

\Rightarrow 52( - 1) - 16c = - 56

\Rightarrow c = \frac{1}{4}

So the other quadratic factor is $x^{2} - x + \frac{1}{4} = \left( x - \frac{1}{2} \right)^{2}$ , giving a repeated root $x = \frac{1}{2}$ .

Comparing the remaining coefficients,

s = 4c - 16b + 52 = 4\left( \frac{1}{4} \right) - 16( - 1) + 52 = 69

t = 52c = 52\left( \frac{1}{4} \right) = 13

To check your answer: with

s = 69

,

t = 13

, substituting

x = \frac{1}{2}

into

4x^{4} - 20x^{3} + 69x^{2} - 56x + 13

gives

\frac{1}{4} - \frac{5}{2} + \frac{69}{4} - 28 + 13 = 0

. ✓

$\therefore$ the other roots are $x = 2 + 3i$ and $x = \frac{1}{2}$ (repeated), with $s = 69$ and $t = 13$ .

Part (b)(i)

3 marks

Approach

A non-calculator method means we factorise $w^{3} - 27$ using the known root $w = 3$ , rather than solving $w^{3} = 27$ directly.

Since $w = 3$ is a root of $w^{3} - 27 = 0$ , $(w - 3)$ is a factor.

w^{3} - 27 = (w - 3)\left( w^{2} + 3w + 9 \right)

The other values of $w$ satisfy $w^{2} + 3w + 9 = 0$ . By the quadratic formula,

w = \frac{- 3 \pm \sqrt{9 - 36}}{2}

= \frac{- 3 \pm \sqrt{- 27}}{2}

= \frac{- 3 \pm 3i\sqrt{3}}{2}

$\therefore$ the other values are $w = - \frac{3}{2} + \frac{3\sqrt{3}}{2}i$ and $w = - \frac{3}{2} - \frac{3\sqrt{3}}{2}i$ .

Part (b)(ii)

2 marks

For each value of $w$ , $|w| = \sqrt{\left( \frac{3}{2} \right)^{2} + \left( \frac{3\sqrt{3}}{2} \right)^{2}} = \sqrt{\frac{9}{4} + \frac{27}{4}} = 3$ , consistent with all cube roots of $27$ having modulus $27^{1/3} = 3$ .

Since $\tan\left( \arg w \right) = \frac{3\sqrt{3}/2}{- 3/2} = - \sqrt{3}$ and the point lies in the second quadrant (for the $\text{+}$ case) or third quadrant (for the $\text{-}$ case), the arguments are $\frac{2\pi}{3}$ and $- \frac{2\pi}{3}$ .

w = 3,\ \ w = 3\left( \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} \right),\ \ w = 3\left( \cos\frac{2\pi}{3} - i\sin\frac{2\pi}{3} \right)

Note: the three points lie on a circle of radius $3$ centred at the origin, spaced $\frac{2\pi}{3}$ apart, since they are the three cube roots of $27$ .

Part (b)(iii)

2 marks

Approach

Rather than adding and multiplying the surd forms directly, we use the fact that the three values of $w$ are the roots of $w^{3} - 27 = 0$ , so their sum and product follow from the coefficients.

Since $w^{3} + 0w^{2} + 0w - 27 = 0$ has roots $3$ , $- \frac{3}{2} + \frac{3\sqrt{3}}{2}i$ , $- \frac{3}{2} - \frac{3\sqrt{3}}{2}i$ ,

\text{Sum of roots} = - \frac{0}{1} = 0

\text{Product of roots} = - \frac{- 27}{1} = 27

To check your answer:

3\left( - \frac{3}{2} + \frac{3\sqrt{3}}{2}i \right)\left( - \frac{3}{2} - \frac{3\sqrt{3}}{2}i \right) = 3\left( \frac{9}{4} + \frac{27}{4} \right) = 3(9) = 27

. ✓

$\therefore$ the sum of all possible values of $w$ is $0$ and the product is $27$ .

2018 H2 Maths Paper 2, Question 2

Part (a)

Part (b)(i)

Part (b)(ii)

Part (b)(iii)

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