Higher November 2023 Paper 4 Q9
9 Here are two pieces of work.
Each shows a question and an incorrect solution.
For each part, describe the error made and write out a correct solution.
(a)
| Question: Factorise. \(x^2 + x - 20\) Solution: \((x + 4)(x - 5)\) |
The error is ……
A correct solution is …… [2]
(b)
| Question: Solve. \(4x + 5 = x + 2\) Solution: \(4x + 5 = x + 2\) \(3x + 5 = 2\) \(3x = 5 - 2\) \(3x = 3\) \(x = 1\) |
The error is ……
A correct solution is …… [2]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Expanding the brackets gives \(-x\) oe | 1 | Accept any correct response, look for answers in the working space (see appendix) | |
| \((x - 4)(x + 5)\) | 1 | ||
Appendix: exemplar responses for Q9(a)
| Response | Mark |
|---|---|
| Expanding brackets gives \(-x\) | 1 |
| The signs are the wrong way round (condone inverted) | 1 |
| The minus sign is on the wrong one | 1 |
| They are the wrong factors | 0 |
| The answer is \((x - 4)(x + 5)\) | 0 |
| The answer is 4 or – 5 | 0 |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| The third line should be 2 – 5 | 1 | Accept any correct response, look for answers in the working space (see appendix) | |
| \(3x = 2 - 5\) \(3x = -3\) \(x = -1\) | 1 | ||
Appendix: exemplar responses for Q9(b)
| Response | Mark |
|---|---|
| [The third line] should be 2 – 5 | 1 |
| She did 5 – 2 it should be 2 – 5 | 1 |
| It should be – 5 not – 2 | 1 |
| 5 should be subtracted by 2 | 0 |
| The error is in \(3x = 5 - 2\) | 0 |