Higher June 2019 Paper 3 Q18
18
(a) Write \(\quad x(3x - 9) = 4 \quad\) in the form \(\quad ax^2 + bx + c = 0 \quad\) where \(a\), \(b\) and \(c\) are integers. [1 mark]
(b) Solve \(\quad x(3x - 9) = 4\)
Give your answers to 2 decimal places. [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3x^2 - 9x - 4 = 0\) or \(-3x^2 + 9x + 4 = 0\) | B1 | must see = 0 on answer line |
Additional guidance
| Do not accept \(x9\) or \(9 \times x\) for \(9x\) | |
| \(3x^2 + {-9x} + {-4} = 0\) | B1 |
| \(3x^2 - {+9x} - {+4} = 0\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{-{-9} \pm \sqrt{(-9)^2 - 4 \times 3 \times -4}}{2 \times 3}\) or \(\dfrac{9 \pm \sqrt{129}}{6}\) or \(\left(x - \dfrac{3}{2}\right)^2 - \dfrac{9}{4} = \dfrac{4}{3}\) or \(\dfrac{3}{2} \pm \sqrt{\dfrac{43}{12}}\) or 3.392… or 3.393 or \(-0.392…\) or \(-0.393\) | M1 | oe correct or ft their 3-term quadratic seen |
| 3.39 and \(-0.39\) | A1ft | correct or ft their 3-term quadratic seen ft answers must be rounded to 2 dp |
Additional guidance
| The word ‘and’ does not need to be seen to award A mark | |
| Full fraction line, correct full square root, \(\pm\) and \((-9)^2\) or \(9^2\) must be seen to award M1 but can be recovered by sight of correct solution(s) | |
| \(3x^2 - 9x + 4 = 0\) in 18(a) \(\dfrac{9 \pm \sqrt{33}}{6}\) or \(\dfrac{3}{2} \pm \sqrt{\dfrac{11}{12}}\) or 2.457… or 0.542… 2.46 and 0.54 | M1 A1ft |
| 3.39 and \(-0.39\) on answer line with no incorrect working | M1A1 |
| 2.46 and 0.54 on answer line with no incorrect working | M1A1ft |
| One correct answer with no incorrect working | M1A0 |