Higher November 2018 Paper 2 Q18
18. Solve the following simultaneous equations.
\[\begin{aligned}y &= 4x^2 + 5x - 7\\ y &= 3x + 2\end{aligned}\]Use an algebraic method and give your answers correct to 2 decimal places. [6]
| Scheme | Marks | Comments |
|---|---|---|
| \(3x + 2 = 4x^2 + 5x - 7\) | M1 | |
| \(4x^2 + 2x - 9 = 0\) | A1 | Must be equated to zero. ‘=0’ may be implied in further work to solve, if no further work and not ‘=0’ then A0 |
| \(x = \dfrac{-2 \pm \sqrt{2^2 - 4\times4\times-9}}{2\times4}\) | m1 | FT provided their quadratic does not factorise and equivalent level of difficulty Use of correct quadratic formula, allow 1 slip in substitution (not a slip with the formula) If completing the square used award m1 for sight of \(4(x + 1/4)^2 \pm \ldots\) |
| \(x = \dfrac{-2 \pm \sqrt{148}}{8}\) | A1 | |
| \(x = 1.27069\ldots\) or \(x = 1.27\) and \(x = -1.77069\ldots\) or \(x = -1.77\) | A1 | |
| \(x = 1.27\) with \(y = 5.81\) (or \(y = 5.80\)) and \(x = -1.77\) with \(y = -3.31\) (or \(y = -3.32\)) | A1 | FT provided M1, m1 previously awarded using their values of \(x\) in \(3x + 2\) or equivalent to find \(y\)-values to 2 d.p. Accept answers given as coordinates |
| (6) |
Notes
Alternative using \(x = (y - 2)/3\)
| \(y = 4\dfrac{(y - 2)^2}{3^2} + 5\dfrac{(y - 2)}{3} - 7\) or equivalent | M1 |
| \(4y^2 - 10y - 77 = 0\) or equivalent, must equate to zero | A1 |
| \(y = \{10 \pm \sqrt{(-10)^2 - 4\times4\times-77}\}/2\times4\) or equivalent, allow 1 slip in substitution | m1 |
| \(y = (10 \pm \sqrt{1332})/8\) or equivalent | A1 |
| \(y = -3.312\ldots\) or \(y = -3.31\) and \(y = 5.812\ldots\) or \(y = 5.81\) | A1 |
| \(x = 1.27,\ y = 5.81\) with \(x = -1.77,\ y = -3.31\) | A1 |
FT to final A1, provided M1, m1 previously awarded using their values of \(y\) in \((y - 2)/3\) or equivalent to find \(x\)-values to 2 d.p.