Higher January 2020 Paper 1R Q6
6 Solve \(\;x^2 - 5x - 36 = 0\)
Show clear algebraic working.
(3)
| Scheme | Marks |
|---|---|
\((x \pm 9)(x \pm 4)\) or \(\dfrac{-(-5) \pm \sqrt{(-5)^2 - 4 \times 1 \times (-36)}}{2 \times 1}\) or \(\dfrac{5 \pm \sqrt{25 + 144}}{2}\) | M1 |
\((x - 9)(x + 4)\) or \(\dfrac{5 \pm \sqrt{169}}{2}\) or \(\dfrac{5 \pm 13}{2}\) | M1 |
| Working required Answer: 9, −4 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: or \((x + a)(x + b)\) where \(ab = -36\) or \(a + b = -5\)
OR correct substitution into quadratic formula (condone one sign error in \(a\), \(b\) or \(c\))
(if + rather than ± shown then award M1 only unless recovered with answers)
M1: or \(\dfrac{5 \pm \sqrt{169}}{2}\) or \(\dfrac{5 \pm 13}{2}\)
A1: dep on at least M1