Higher January 2022 Paper 2R Q18
18 The line with equation \(2y = x + 1\) intersects the curve with equation \(3y^2 + 7y + 16 = x^2 - x\) at the points \(A\) and \(B\)
Find the coordinates of \(A\) and the coordinates of \(B\)
Show clear algebraic working.
(5)
| Scheme | Marks |
|---|---|
\(3y^2 + 7y + 16 = (2y - 1)^2 - (2y - 1)\) or \(3\left(\dfrac{x + 1}{2}\right)^2 + 7\left(\dfrac{x + 1}{2}\right) + 16 = x^2 - x\) | M1 |
| E.g. \(y^2 - 13y - 14\ (= 0)\) oe \(y^2 - 13y = 14\) or E.g. \(x^2 - 24x - 81\ (= 0)\) oe \(x^2 - 24x = 81\) | A1 |
E.g. \((y - 14)(y + 1)\ (= 0)\) or \((y =)\ \dfrac{-(-13) \pm \sqrt{(-13)^2 - 4 \times 1 \times -14}}{2}\) or \(\left(y - \dfrac{13}{2}\right)^2 - \left(\dfrac{13}{2}\right)^2 = 14\) oe or E.g. \((x + 3)(x - 27)\ (= 0)\) or \((x =)\ \dfrac{-(-24) \pm \sqrt{(-24)^2 - 4 \times 1 \times -81}}{2}\) or \(\left(x - \dfrac{24}{2}\right)^2 - \left(\dfrac{24}{2}\right)^2 = 81\) oe | M1 |
\((x =)\ 2 \times \text{‘}{14}\text{’} - 1\) and \(2 \times \text{‘}{-1}\text{’} - 1\) or \((y =)\ \dfrac{\text{‘}{27}\text{’} + 1}{2}\) and \(\dfrac{\text{‘}{-3}\text{’} + 1}{2}\) oe | M1 |
| Working required Answer: \((27, 14)\) and \((-3, -1)\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: substitution of linear equation into quadratic.
A1: (dep on M1) writing the correct quadratic expression in form \(ax^2 + bx + c\ (= 0)\)
allow \(ax^2 + bx = c\)
M1: (dep on M1) for the first stage to solve their 3-term quadratic equation (allow one sign error and some simplification – allow as far as \(\dfrac{13 \pm \sqrt{169 + 56}}{2}\) or \(\dfrac{24 \pm \sqrt{576 + 324}}{2}\)
or eg \(\left(x - \dfrac{24}{2}\right)^2 - 225\) oe
M1: (dep on previous M1)
may be implied by values of \(y\) or \(x\) that are consistent with a correct substitution.
A1: for both solutions dep on M2
Must be paired correctly.
accept \(x = 27\), \(y = 14\) and \(x = -3\), \(y = -1\)