Higher January 2020 Paper 2 Q7
7
(a) Solve the inequality \(\;2x + 7 \gt 4\) (2)
(b) Solve \(\;x^2 - 3x - 40 = 0\)
Show clear algebraic working. (3)
Show clear algebraic working. (3)
| Scheme | Marks |
|---|---|
| \(2x \gt 4 - 7\) or \(x + 3.5 \gt 2\) | M1 |
| \(x \gt -1.5\) | A1 |
| (2) |
Notes
M1: For a correct first step allow \(2x = 4 - 7\) or \(x + 3.5 = 2\) or an answer of \(x = -1.5\) or \(x \lt -1.5\) or −1.5
A1: for \(x \gt -1.5\) oe
| Scheme | Marks |
|---|---|
\((x \pm 8)(x \pm 5)\) or \(\dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 1 \times (-40)}}{2 \times 1}\) or \(\dfrac{3 \pm \sqrt{9 + 160}}{2}\) | M1 |
\((x - 8)(x + 5)\) or \(\dfrac{3 \pm \sqrt{169}}{2}\) or \(\dfrac{3 \pm 13}{2}\) | M1 |
| Working required Answer: 8, −5 | A1 |
| (3) | |
| (5 marks) |
Notes
M1: or \((x + a)(x + b)\) where \(ab = -40\) or \(a + b = -5\)
OR correct substitution into quadratic formula (condone one sign error in \(a\), \(b\) or \(c\) and missing brackets)
(if + rather than ± shown then award M1 only unless recovered with answers)
M1: \(\dfrac{3 \pm \sqrt{169}}{2}\) or \(\dfrac{3 \pm 13}{2}\)
A1: dep on at least M1 for correct values