Higher June 2025 Unit 2 Q15
15. Triangle \(PQR\) is shown below.
The length of \(PQ\) is \((2x + 1)\) cm.
The length of \(QR\) is \(x\) cm.
The length of \(PR\) is 6 cm.
The size of \(P\hat{R}Q\) is 60°.
Diagram not drawn to scale
(a) Use the cosine rule to show that \(3x^2 + 10x - 35 = 0\). [3]
(b) Solve the equation\[3x^2 + 10x - 35 = 0.\]You must use the quadratic formula and show all your working.
Give your answers correct to 2 decimal places. [3]
Give your answers correct to 2 decimal places. [3]
(c) Evaluate the length of \(PQ\).
You must justify any decision that you make. [2]
You must justify any decision that you make. [2]
From the formula list: In any triangle \(ABC\): Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\); The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| \((2x + 1)^2 = x^2 + 6^2 - 2 \times x \times 6 \times \cos 60^\circ\) or equivalent | B1 | Radians: cos 60 = −0·952, gradians: cos 60 = 0·587 |
| \(4x^2 + 2x + 2x + 1\) OR \(4x^2 + 4x + 1\) | B1 | Correct expansion of \((2x + 1)^2\) |
| \(4x^2 + 2x + 2x + 1 = x^2 + 36 - 2 \times x \times 6 \times 0{\cdot}5\) OR \(4x^2 + 4x + 1 = x^2 + 36 - 6x\) or equivalent leading to: \(3x^2 + 10x - 35 = 0\) | B1 | Correct solution only. Correct collection of terms. Must be convincing. |
| Scheme | Marks | Comments |
|---|---|---|
| Solution may be seen in part (a). If seen in both, the work in the answer space for part (b) takes precedence. | ||
| (\(x\) =) \(\dfrac{-10 \pm \sqrt{10^2 - 4(3)(-35)}}{2(3)}\) | M1 | Evidence of this substitution into the formula must be seen for M1, otherwise award M0A0A0. Allow one slip in substitution for M1 only, but must be correct formula. |
| (\(x\) =) \(\dfrac{-10 \pm \sqrt{520}}{6}\) \(\left(= \dfrac{-5 \pm \sqrt{130}}{3}\right)\) | A1 | Can be implied from at least one correct value of x evaluated, provided M1 awarded. Sight of this alone is insufficient evidence of substitution into the formula. |
| (\(x\) =) –5·47 AND 2·13 | A1 | Both solutions required (but ISW for discarding the negative root). Award M0A0A0 for use of trial and improvement. An unsupported answer gains zero marks. |
| Scheme | Marks | Comments |
|---|---|---|
| Solution may be seen in part (a) or (b). If seen in both, the work in the answer space for part (c) takes precedence. | ||
| Unique answer of: (\(PQ\) = 2×2·13 + 1 =) 5·26 (cm) | B1 | F.T. ‘their derived \(x\)’ provided one positive and one negative solution. Use of \(x\) = 2·1339… leads to an answer of 5·2678…. or 5·27 (cm). Allow 5·3 (cm) Award B0 if both a positive and negative answer is offered for PQ, unless corrected in the explanation. |
| Length cannot be negative / must be positive. | E1 | Accept any valid explanation, e.g. ‘(Length cannot be negative as) x cannot be negative’ e.g. \(x \gt 0\) (or allow \(x \gt -\tfrac{1}{2}\)). Ignore additional, spurious or invalid comments. |