A2 June 2022 Paper 1 Q10

10 The equation

\(4x^4 + 16x^3 + ax^2 + bx + 6 = 0\),

where \(a\) and \(b\) are real, has roots \(\alpha\), \(\dfrac{2}{\alpha}\), \(\beta\) and \(3\beta\).

(a) Given that \(\beta \lt 0\), determine all 4 roots. [6]

(b) Determine the values of \(a\) and \(b\). [4]

Mark scheme (a)

Scheme	Marks	AO
\(\alpha \times \dfrac{2}{\alpha} \times \beta \times 3\beta = \dfrac{6}{4}\)	M1	3.1a
\(\Rightarrow \beta = -\dfrac{1}{2}\)	A1	1.1
\(\alpha + \dfrac{2}{\alpha} + \beta + 3\beta = -\dfrac{16}{4}\)	M1	3.1a
\(\Rightarrow \alpha^2 + 2\alpha + 2 = 0\)	A1	1.1
\(\alpha = \dfrac{-2 \pm \sqrt{4 - 8}}{2} = -1 + i,\ -1 - i\)	M1	1.1
so roots are \(-1 + i,\ -1 - i,\ -\frac{1}{2},\ -\frac{3}{2}\)	A1	3.2a
	[6]

M1: Solving their quadratic to find \(\alpha\)

Mark scheme (b)

Scheme	Marks	AO
\((-1 + i)(-1 - i) + (-1 + i)\left(-\frac{1}{2}\right) + (-1 + i)\left(-\frac{3}{2}\right)\) \(+ (-1 - i)\left(-\frac{1}{2}\right) + (-1 - i)\left(-\frac{3}{2}\right) + \left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = \dfrac{a}{4}\)	M1	1.1
\(\Rightarrow a = 27\)	A1	1.1
\((-1 + i)(-1 - i)\left(-\frac{1}{2}\right) + (-1 + i)(-1 - i)\left(-\frac{3}{2}\right)\) \(+ (-1 + i)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) + (-1 - i)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right) = -\dfrac{b}{4}\)	M1	1.1
\(\Rightarrow b = 22\)	A1	1.1
	[4]

M1: Allow in terms of \(\alpha\) and \(\beta\), no missing terms

Scheme	Marks
\((x + 1 + i)(x + 1 - i)(2x + 1)(2x + 3) = 0\)	M1
\(\Rightarrow \left(x^2 + 2x + 2\right)\left(4x^2 + 8x + 3\right) = 0\)	M1
\(\Rightarrow 4x^4 + 16x^3 + 27x^2 + 22x + 6 = 0\)	A1
\(\Rightarrow a = 27,\ b = 22\)	A1

Scheme	Marks
\(f\left(-\dfrac{3}{2}\right) = \dfrac{81}{4} - 54 + \dfrac{9a}{4} - \dfrac{3b}{2} + 6 = 0\)	M1
\(f\left(-\dfrac{1}{2}\right) = \dfrac{1}{4} - 2 + \dfrac{a}{4} - \dfrac{b}{2} + 6 = 0\)	M1
\(a = 27\)	A1
\(b = 22\)	A1
	[4]