4MA0 Higher June 2014 Paper 4R Q16

EdexcelHigherOld spec8 marksForming Equations Solving Quadratics

16 Here is a hexagon.

Hexagon with all corners right angles; sides labelled 4, 3x, (4x - 3) and (x + 1)

Diagram NOT accurately drawn

In the diagram, all the measurements are in centimetres.
All the corners are right angles.

The area of the hexagon is 40 cm²

(a) Show that \(\;4x^2 + 9x - 47 = 0\) (3)

(b) Solve \(\;4x^2 + 9x - 47 = 0\)
Show your working clearly.
Give your solutions correct to 3 significant figures. (3)

Mark scheme (a)

Scheme	Marks
Eg. Area = \((4x - 3)(x + 1) + [3x - (x + 1)] \times 4\) OR Area = \(4 \times 3x + (x + 1)(4x - 3 - 4)\) OR Area = \((4x - 3) \times 3x - (4x - 3 - 4)(3x - (x + 1))\)	M1
	A1
Answer given	A1
	(3)

Notes

M1: for a complete correct expression for area of hexagon

A1: for all brackets correctly expanded

A1: for convincing progression to given equation

Mark scheme (b)

Scheme	Marks
\(\dfrac{-9 \pm \sqrt{9^2 - 4 \times 4 \times (-47)}}{2 \times 4}\)	M1
\(\sqrt{833}\) or \(\sqrt{81 + 752}\) or \(7\sqrt{17}\) or 28.8…..	M1
2.48, − 4.73	A1
	(3)

Notes

M1: for correct substitution; condone one sign error; brackets not necessary;
condone + instead of ± in formula
There may be partial evaluation – if so, this must be correct

M1: (independent) for correct simplification of discriminant (if evaluated, at least 3sf rounded or truncated)

A1: awrt 2.48 and −4.73
NB. If negative solution is discarded (or omitted at any stage) then full marks can still be obtained
Award 3 marks if first M1 scored and answer correct

Alternative

Scheme	Marks
\(x^2 + \dfrac{9}{4}x - \dfrac{47}{4} = 0\) \(\left(x + \dfrac{9}{8}\right)^2 - \dfrac{81}{64} - \dfrac{47}{4} = 0\)	M1
\(\left(x + \dfrac{9}{8}\right) = \pm\sqrt{\dfrac{833}{64}}\)	M1
2.48, − 4.73	A1

Notes

M1: for \(\left(x + \dfrac{9}{8}\right)^2\) oe

M1: for \(\left(x + \dfrac{9}{8}\right) = \pm\sqrt{\dfrac{833}{64}}\) oe

A1: awrt 2.48 and −4.73
NB. If negative solution is discarded (or omitted) then full marks can still be obtained
Award 3 marks if first M1 scored and answer correct

The mark scheme states that for this question the correct answer alone does not imply a correct method.

Scheme	Marks
3 ×“2.48”.... (=7.44…) or 4 × “2.48” − 3 (=6.93…)	M1
7.45	A1
	(2)
	(8 marks)