4400 Higher June 2010 Paper 3 Q21

21 Solve the simultaneous equations

\[\begin{aligned} y &= 2x^2 \\ y &= 3x + 14 \end{aligned}\]

(5)

Mark scheme

Scheme	Marks
\(2x^2 = 3x + 14\) May be implied by second M1	M1
\(2x^2 - 3x - 14\ (= 0)\)	M1
\((2x - 7)(x + 2)\ (= 0)\) or \(\dfrac{3 \pm \sqrt{121}}{4}\) or \(\dfrac{3}{4} \pm \dfrac{\sqrt{121}}{4}\)	M1
\(x = \frac{7}{2}\), \(x = -2\)	A1
\(x = \frac{7}{2}\), \(y = \frac{49}{2}\) \(x = -2\), \(y = 8\)	A1
	(5)
	(5 marks)

M1: \(y = 2\left(\dfrac{y - 14}{3}\right)^2\)

M1: \(2y^2 - 65y + 392 = 0\)

M1: \((2y - 49)(y - 8)\ (= 0)\)

or \(\dfrac{65 \pm \sqrt{1089}}{4}\)

or \(\dfrac{65}{4} \pm \dfrac{\sqrt{1089}}{4}\)

A1: dep on all method marks
\(y = \frac{49}{2}\), \(y = 8\)

A1: dep on all method marks
\(x = \frac{7}{2}\), \(y = \frac{49}{2}\)
\(x = -2\), \(y = 8\)