Higher June 2019 Paper 1 Q17
17 Given that
\[x^2 : (3x + 5) = 1 : 2\]find the possible values of \(x\). (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(-1\), 2.5 | P1 | for process to form an equation, eg \(\dfrac{x^2}{3x + 5} = \dfrac{1}{2}\) or \(2x^2 = 3x + 5\) |
| P1 | for writing in a suitable form ready for solution, eg \(2x^2 - 3x - 5\ (= 0)\) or \(-2x^2 + 3x + 5\ (= 0)\) | |
| P1 | (dep 1st P1) for process to solve quadratic equation of form \(ax^2 + bx + c\ (= 0)\) eg \((2x - 5)(x + 1)\ (= 0)\) or \(\dfrac{--3 \pm \sqrt{(-3)^2 - 4 \times 2 \times -5}}{2 \times 2}\) | |
| A1 | for \(-1\), 2.5 oe |