Higher June 2018 Unit 2 Q17
17. Solve the equation \((2x + 5)(3x - 11) = 7\).
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 |
|---|---|---|
| \(6x^2 - 22x + 15x - 55\) (= 7) | B1 | |
| \(6x^2 - 7x - 62 = 0\) | B1 | CAO. '= 0' may be implied in further working. |
| \((x =)\ \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 6 \times (-62)}}{2 \times 6}\) | M1 | FT ‘their derived quadratic equation’ set to zero and of equivalent level of difficulty (\(a\), \(b\) and \(c\) are non-zero). Allow one slip in substitution, but must be correct formula. |
| \(= \dfrac{7 \pm \sqrt{1537}}{12}\) | A1 | If one slip seen award A0. |
| x = 3.85 AND x = –2.68 (answers to 2dp) | A1 | CAO for 'their equation'. Note: no marks to be awarded for 1 correct solution from trial and improvement. |