Forming Equations: WJEC GCSE Higher November 2017 Unit 1 Q12

Higher November 2017 Unit 1 Q12

WJECHigherCurrent spec9 marksForming Equations Solving Quadratics

12. Two different squares are constructed.
The side length of the smaller square is \(x\) cm.
The side length of the larger square is 3 cm longer than the side length of the smaller square.
The combined area of the two squares is 22·5 cm².

(a) Show that \(4x^2 + 12x - 27 = 0\). [4]

(b) Find the dimensions of each of the squares.
Do not use a trial and improvement method.
You must show all your working and justify any decision that you make. [5]

Side length of smaller square = …………………… cm
Side length of larger square = …………………… cm

From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Mark scheme (a)

Scheme	Marks	Comments
(Total area =) \(x^2 + (x + 3)^2\) or equivalent	B1	Allow award of B1 if brackets are omitted
\(x^2 + x^2 + 3x + 3x + 9\)	M1
\(2x^2 + 6x + 9 = 22.5\)	A1	F.T. for equivalent difficulty i.e. from \(x^2 + (ax + b)^2\) with a,b ≠ 0.
\(4x^2 + 12x - 27 = 0\)	A1	Equating to zero and doubling. Must be convincing.

Mark scheme (b)

Scheme	Marks	Comments
\((2x - 3)(2x + 9) = 0\)	B2	B1 for \((2x \ldots 3)(2x \ldots 9)\)
\(x = 3/2\) [or \(x = -9/2\)]	B1	FT from ‘their two brackets’. (If both F.T. solutions are of the same sign, then both are required for this B1.) Ignore presence or absence of x = –9/2.
(Dimensions are) 3/2 (cm) and (3/2 + 3 =) 9/2 (cm)	B1	F.T. ‘their derived \(x\)’.
Explanation that \(x\) cannot be – 9/2 (cm) because a length cannot be negative (or must be positive).	E1	F.T. provided one solution is positive and the other is negative.

Alternative method (using quadratic formula):

Scheme	Marks	Comments
\(x = [-12 \pm \sqrt{(12^2 - 4 \times 4 \times -27)}\,] / (2 \times 4)\)	M1	Allow one error, in sign or substitution, but not in the formula.
\(x = [-12 \pm \sqrt{576}\,] / 8\)	A1	C.A.O.
\(x = 3/2\) [or \(x = -9/2\)]	A1	C.A.O.