Higher November 2018 Unit 2 Q18
18. Solve the equation \(\dfrac{7x + 1}{5x + 2} = \dfrac{1}{x + 3}\) .
Give your answers correct to 2 decimal places.
You must show all your working. [6]
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 |
|---|---|---|
| \((7x + 1)(x + 3) = 5x + 2\) | B1 | |
| Sight of \(7x^2 + 21x + x + 3\) | B1 | Or equivalent. |
| \(7x^2 + 17x + 1 = 0\) | B1 | '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 7, b = 17,\) \(c = 1\) used in the quadratic formula. FT from B1B0 from one error only. |
| \(x = \dfrac{-17 \pm \sqrt{17^2 - 4 \times 7 \times 1}}{2 \times 7}\) | M1 | This substitution into the formula must be seen for M1. FT ‘their derived quadratic equation’ of equivalent difficulty (\(a\), \(b\) and \(c\) must be non-zero). Award M1A0A0 for only one slip in substitution, but must be correct formula. |
| \(x = \dfrac{-17 \pm \sqrt{261}}{14}\) | A1 | Can be implied from at least one correct value of \(x\) evaluated. |
| \(x = -0.06\) with \(x = -2.37\) (answers to 2dp) | A1 | CAO for their quadratic equation. If trial and improvement used, then award: SC3 for both correct solutions given, correct to 2 decimal places: x = -0.06 with x = -2.37, OR SC2 for both correct solutions given, but correct to 3 (or more) decimal places: x = -0.060(3…) with x = –2.368(2…) Note: no marks to be awarded for 1 correct solution from trial and improvement. |
Corrected by Westie's Workshop
- The board's mark scheme printed “x = –-2.37”. Changed to “x = -2.37” because the double minus sign is a printing slip: the solution is −2.37, as in the line above.
Original: see the board's mark scheme PDF (link above).