June 2023 Paper 1 Q15
15 Solve \(\quad \left(x^{\frac{1}{2}} - x^{\frac{3}{2}}\right)^2 = x^2 + x\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\) | M1 | allow error in 1 term but must not have missing terms (could be seen in a grid) if they go straight to \(x - 2x^2 + x^3\) then M1 if they go straight to \(x - kx^2 + x^3\) where \(k\) is not 2 then this could be more than one error so M0 |
| \(x^3 - 3x^2 = 0\) or \(x^3 = 3x^2\) | M1dep | no ft |
| \(x^2(x - 3)(= 0)\) | M1 | factorising their equation fully and correctly (must have \(x^3\) plus/minus other term(s) in \(x^n\) where n\(\neq\)0). |
| (\(x =\)) 0 and (\(x =\)) 3 | A1 | condone \(x = 0\) \(x = 0\) \(x = 3\) |
Additional guidance
| First M mark could be done in root form but would still need to end up with \(x - x^2 - x^2 + x^3 \quad (= x^2 + x)\) for M1 | |
| Candidates may divide by \(x\) at some point during the method. If they do this and state that \(x = 0\) is a solution at this point, the method can be continued to a correct outcome for full marks or \(x^2 - 3x = 0\) or \(x^2 = 3x\) would be M1dep and \(x(x - 3)(= 0)\) would be the third M mark eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) with \(x = 0\) \(x^2 - 3x = 0\) If they do not state \(x = 0\) is a solution and continue to a correct solution then award full marks but if it’s not stated and they don’t get to the correct solution then no further marks available as soon as the division is done eg \(x - x^2 - x^2 + x^3 = x^2 + x\) \(1 - x - x + x^2 = x + 1\) (no mention of \(x = 0\) being a possible solution) \(x^2 - 3x = 0\) | M1 M1dep M1 M0dep |
| If no other marks awarded then \(x = 0\) or \(x = 3\) seen on answer line (or the last thing they have written) | SC1 |
| \(x = 0\) and \(x = 3\) not from incorrect working | M3 A1 |