Higher November 2019 Unit 1 Q8
8.
(a) Factorise \(x^2 + 4x - 21\). Hence, solve \(x^2 + 4x - 21 = 0\). [3]
(b) Solve the equation \(\dfrac{2x-3}{5} + \dfrac{4x+5}{2} = \dfrac{11}{2}\). [4]
| Scheme | Marks | Comments |
|---|---|---|
| \((x + 7)(x - 3)\) | B2 | B1 for \((x \;...\; 7)(x \;...\; 3)\). |
| (\(x\) =) −7 AND (\(x\) =) 3 | B1 | Strict F.T. from their brackets. Allow the following. B2 for \(x + 7\) (=0) AND \(x - 3\) (=0) (B1) (\(x\) =) −7 AND (\(x\) =) 3 (B1) B1 for \(x - 7\) (=0) AND \(x + 3\) (=0) (B0) (\(x\) =) 7 AND (\(x\) =) −3 (B1) FT B1 if only (\(x\) =) −7 AND (\(x\) =) 3 seen. (B1) |
| Scheme | Marks | Comments |
|---|---|---|
| FT until 2nd error. | ||
| Correct method for clearing all three fractions. | M1 | May be seen in stages. |
| Accurate clearing of fractions AND expansion of brackets on lhs. | A1 | Allow if all over a common denominator. May be seen in stages |
| \(24x = 36\) or equivalent. | A1 | For collection of terms. |
| \(x = \dfrac{36}{24}\) or equivalent | A1 | FT from ‘their \(ax = b\)’ ONLY if M1 gained AND no more than one previous error. If no marks, allow SC1 for sight of \(\dfrac{2(2x - 3) + 5(4x + 5)}{(10)}\) or equivalent. If FT answer is a whole number then it must be shown as an integer. Allow a correct embedded answer of 1·5 or 1½ BUT penalise −1 if followed by \(x \ne 1{\cdot}5\) or 1½. Note : An answer of 1·5 that is found without gaining M1 OR that is not embedded is zero marks. |