Higher June 2025 Paper 6 Q9
9
(a) Solve.
\(8x - 11 = 6x + 3\) [3]
(b) Solve by factorising.
\(x^2 + 15x - 16 = 0\) [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 7 | 3 | M2 for \(8x - 6x = 3 + 11\) oe OR M1 for \(2x - 11 = 3\) or for \(2x = k\) or for \(8x = 14 + 6x\) or for \(kx = 14\) M1 for \(x = \frac{b}{a}\) FT their \(ax = b\) seen | Embedded 7 scores M2 e.g. 8 × 7 – 11 = 6 × 7 + 3 [= 45] identified as final working or written on answer line eg \(2x = -8\) and answer line –4 with no other working scores M1 and M1FT |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x + 16)(x - 1)\) | M2 | Condone final bracket missing Accept \(x + 16 = 0\) and \(x - 1 = 0\) for M2 | |
| M1 for \((x + a)(x + b)\) where \(ab = -16\) or \(a + b = 15\) or \(x(x + 16) - [1](x + 16)\) or \(x(x - 1) + 16(x - 1)\) | Allow correctly completed grid for M1 | ||
| –16 and 1 | B1FT | For correct solutions from their quadratic factors If 0 scored, instead award SC1 for answers ±16 and ±1 | \((x - 16)(x + 1)\) with answers –16 and 1 scores M1B0 |