Higher June 2023 Paper 3 Q19
19 Two integers have a difference of 6
The integers are multiplied together.
9 is then added.
Prove algebraically that the result is always a square number. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Creates an algebraic product in the form \((x + a)(x + b)\) where there is a difference of 6 between \(a\) and \(b\) | M1 | accept any letter for \(x\) eg \(x(x + 6)\) or \(x^2 + 6x\) or \(x(x - 6)\) or \(x^2 - 6x\) |
| Correctly expands their product, adds 9 and simplifies to a quadratic expression | M1dep | eg \(x^2 + 6x + 9\) or \(x^2 - 6x + 9\) |
| Correctly factorises their quadratic expression to the form \((x + c)^2\) with M2 awarded | A1 | eg \((x + 3)^2\) or \((x - 3)^2\) |
Additional guidance
| Trialling integers scores no marks, but ignore any testing of values alongside correct algebra | |
| Ignore any further work or attempts to solve after correct answer seen | |
| Missing brackets may be recovered eg \(x \times x + 6\) followed by \(x^2 + 6x + 9\) | M1M1 |
| \((x + 3)(x + 3)\) without \((x + 3)^2\) seen does not score the A mark | |
| \((x - 2)(x - 8)\) | M1 |
| \(x^2 - 2x - 8x + 16 + 9 = x^2 - 10x + 25\) | M1 |
| \((x - 5)^2\) | A1 |