Higher June 2017 Unit 2 Q18
18. Solve the equation \(x = \dfrac{7}{5x - 3}\).
Give your answers correct to 2 decimal places. [5]
From the formula list: The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
| Scheme | Marks | Comments |
|---|---|---|
| \(x(5x - 3) = 7\) OR \(7 = x(5x - 3)\) OR \(5x^2 - 3x = 7\) OR \(7 = 5x^2 - 3x\) | M1 | |
| 5x² –3x –7 = 0 | A1 | '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 5, b = -3,\) \(c = -7\) used in the quadratic formula. |
| \(x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 5 \times (-7)}}{2 \times 5}\) | M1 | FT ‘their quadratic equation’ of equivalent difficulty (3 terms with at least one negative term). Allow one slip in substitution, but must be correct formula. |
| = (3±√149)/10 | A1 | |
| x = 1.52 with x = –0.92 (answers to 2dp) | A1 | CAO for their quadratic equation. If none of the last 3 marks awarded for solving the given equation or the correct quadratic (irrespective if any of the opening two marks awarded), and trial and improvement used, then award: SC3 for both correct solutions given, correct to 2 decimal places: x = 1.52 with x = –0.92, OR SC2 for both correct solutions given, but correct to 3 (or more) decimal places: x = 1.520(6…) with x = –0.920(6…) Note: no marks to be awarded for 1 correct solution from trial and improvement. |