Higher June 2018 Paper 1 Q27
27
(a) Jo wants to work out the solutions of \(\quad x^2 + 3x - 5 = 0\)
She says,
“The solutions cannot be worked out because
\(x^2 + 3x - 5\) does not factorise to \(\;(x + a)(x + b)\;\) where \(a\) and \(b\) are integers.”
Is Jo correct?
Tick a box.
- Yes
- No
Give a reason for your answer. [1 mark]
(b) Without expanding any brackets,
show how to work out the exact solutions of \(\quad 9(x + 3)^2 = 4\)
Give the solutions. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Ticks No and gives valid reason | B1 | eg valid reasons could use formula could complete the square could use \(\dfrac{-3 \pm \sqrt{29}}{2}\) |
Additional guidance
| Any working or solutions shown must be correct | |
| If the quadratic formula is written down it must be correct | |
| Ignore irrelevant non-contradictory statements | |
| Ticks No and ‘There are other methods’ | B1 |
| Ticks No and ‘\(a\) and \(b\) could be decimals’ | B1 |
| Ticks No and ‘She could draw a graph’ | B1 |
| Ticks No and ‘All quadratic equations can be solved (even if the solutions aren’t real numbers)’ | B1 |
| Ticks No and ‘The discriminant is positive’ | B1 |
| Ticks No and ‘Not all quadratics factorise’ | B0 |
| Ticks No and ‘It does factorise’ | B0 |
| Ticks Yes | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \((x + 3)^2 = \dfrac{4}{9}\) or \(\sqrt{9}\,(x + 3) = (\pm)\sqrt{4}\) or \(3(x + 3) = (\pm)2\) or \(\left((x + 3) + \dfrac{2}{3}\right)\left((x + 3) - \dfrac{2}{3}\right)\) | M1 | oe |
| \(x + 3 = \pm\sqrt{\dfrac{4}{9}}\) or \(3x = \pm 2 - 9\) or \(x + 3 = \pm\dfrac{2}{3}\) | M1dep | oe eg (\(x =\)) \(-3 \pm \sqrt{\dfrac{4}{9}}\) (\(x =\)) \(\dfrac{2}{3} - 3\) and (\(x =\)) \(-\dfrac{2}{3} - 3\) |
| \(-\dfrac{7}{3}\) and \(-\dfrac{11}{3}\) with correct working for M1M1 | A1 | allow equivalent fractions or recurring decimals or mixed numbers |
Additional guidance
| For up to M1M1, allow 0.66… or 0.67 for \(\dfrac{2}{3}\) and \(-2.33\)… for \(-\dfrac{7}{3}\) and \(-3.66\)… or \(-3.67\) for \(-\dfrac{11}{3}\) | |
| Answers \(-2.33\)… and \(-3.66\)… or \(-3.67\) with correct working | M1M1A0 |
| (\(x =\)) \(-\dfrac{7}{3}\) and (\(x =\)) \(-\dfrac{11}{3}\) with no correct working | M0M0A0 |
| Do not allow incorrect conversion of correct solutions | M1M1A0 |
| Allow \(3(x + 3) = (\pm)\,2\) followed by \(3x + 9 = (\pm)\,2\) etc as a correct method even though it includes a bracket expansion |