Higher November 2019 Paper 2 Q13
13. Solve the following simultaneous equations.
\[\begin{aligned} y &= 3x^2 + 4x - 7\\ y &= 2x + 5 \end{aligned}\]Use an algebraic method and give your answers correct to 2 decimal places. [6]
| Scheme | Marks | Comments |
|---|---|---|
| \(2x + 5 = 3x^2 + 4x - 7\) | M1 | |
| \(3x^2 + 2x - 12 = 0\) | A1 | Must be equated to zero. ‘=0’ may be implied in further work to solve, if no further work and not ‘=0’ then A0 |
| \(x = \dfrac{-2 \pm \sqrt{2^2 - 4 \times 3 \times -12}}{2 \times 3}\) | 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) |
| \(x = \dfrac{-2 \pm \sqrt{148}}{6}\) or \(x = \dfrac{-1 \pm \sqrt{37}}{3}\) | A1 | |
| \(x = 1.69\ldots\) and \(x = -2.36\ldots\) | A1 | |
| \(x = 1.69\) with \(y = 8.39\) and \(x = -2.36\) with \(y = 0.28\) | A1 | FT provided M1, m1 previously awarded using their values of \(x\) in \(2x + 5\) or equivalent to find \(y\)-values to 2 d.p. Accept answers given as coordinates |
| (6) |
Notes
Alternative using \(x = (y - 5)/2\)
| Scheme | Marks |
|---|---|
| Correct substitution of \(x = (y - 5) \div 2\) | M1 |
| \(3y^2 - 26y + 7 = 0\) | A1 |
| Allow 1 slip in substitution into the quadratic formula | m1 |
| Correct simplification of \(b^2 - 4ac\) | A1 |