Forming Equations: WJEC GCSE Higher June 2023 Unit 1 Q13

Higher June 2023 Unit 1 Q13

13. Triangle \(ABC\) is right-angled.
The area of triangle \(ABC\) is 0·75 m².

Diagram not drawn to scale

(a) Show that \(16x^2 - 8x - 3 = 0\) . [3]

(b)

(i) Solve the equation \(16x^2 - 8x - 3 = 0\) .
You must use an algebraic method. [3]

(ii) Find the length of \(BC\).
You must justify any decision that you make. [1]

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
(Area of triangle =) \(\frac{1}{2} \times 4x \times (2x - 1)\)	M1	Allow award of M1 if brackets omitted. Accept equivalent e.g. using area of rectangle = 1·5.
\(\dfrac{8x^2 - 4x}{2} = \dfrac{3}{4}\) or equivalent	m1	Expanding brackets and equating.
\(16x^2 - 8x - 3 = 0\)	A1	Clearing fractions and equating to zero. Must be convincing.

Mark scheme (b)

Scheme	Marks	Comments
(i) \((4x - 3)(4x + 1)\) [= 0]	M2	Solution may be seen in part (a). If seen in both, the work in the answer space for part (b)(i) takes precedence. M1 for \((4x \ldots 3)(4x \ldots 1)\) M1 for two brackets which multiply to give \(16x^2 - 8x + k\) or \(16x^2 + mx - 3\).
\(x = \frac{3}{4}\) AND \(x = -\frac{1}{4}\)	A1	Strict FT from M1
(ii) (\(BC = 2 \times \frac{3}{4} - 1 =\)) 0·5 (m) AND a valid statement e.g. length cannot be negative; length must be positive.	E1	Solution may be seen in part (a) or (b)(i). If seen in both, the work in the answer space for part (b)(ii) takes precedence. FT ‘their \(x\)’ provided an equivalent decision is required i.e. one value of \(x\) is greater than \(\frac{1}{2}\) and the other is less than \(\frac{1}{2}\).

Scheme	Marks	Comments
(\(x\) =) \(\dfrac{-(-8) \pm \sqrt{(-8)^2 - 4(16)(-3)}}{2(16)}\)	M1	For M1, allow one error, in sign or substitution, but not in formula.
\(x = \dfrac{8 \pm \sqrt{256}}{32}\)	A1
\(x = \frac{3}{4}\) or \(x = -\frac{1}{4}\) (both answers required)	A1

Do not allow a trial and improvement method.