Intermediate November 2019 Unit 1 Q19
19.
(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 FT 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 |
|---|---|---|
| Correct method for clearing all three fractions. | M1 | FT until 2nd error. 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 ≠ 1·5 or 1½. Note : An answer of 1·5 that is found without gaining M1 OR that is not embedded is zero marks. |