Higher November 2020 Paper 2 Q29
29 Solve \(\quad \dfrac{5}{4x + 1} = \dfrac{2x}{x^2 + 3}\)
Give your solutions to 3 significant figures.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(5(x^2 + 3)\) or \(5x^2 + 15\) or \(2x(4x + 1)\) or \(8x^2 + 2x\) | M1 | oe ignore any denominators |
| \(5(x^2 + 3) = 2x(4x + 1)\) or \(5x^2 + 15 = 8x^2 + 2x\) | M1dep | oe allow both sides to have denominator \((4x + 1)(x^2 + 3)\) oe |
| \(3x^2 + 2x - 15\ (= 0)\) | M1dep | oe equation with terms collected eg \(3x^2 + 2x = 15\) no denominator allowed unless recovered in subsequent working |
| \(\dfrac{-2 \pm \sqrt{2^2 - 4 \times 3 \times -15}}{2 \times 3}\) or \(\dfrac{-2 \pm \sqrt{184}}{6}\) or \(-\dfrac{1}{3} \pm \dfrac{1}{3}\sqrt{46}\) or 1.927… and \(-2.594\)… and \(3x^2 + 2x - 15\ (= 0)\) seen | M1 | oe ft their 3-term quadratic allow correct factorisation of their 3-term quadratic |
| 1.93 and \(-2.59\) and \(3x^2 + 2x - 15\ (= 0)\) seen | A1 | oe eg 1.93 and \(-2.59\) with \(3x^2 + 2x = 15\) seen |
Additional guidance
| 1.93 and \(-2.59\) and \(3x^2 + 2x - 15\ (= 0)\) not seen | Zero |
| 1.927… and \(-2.594\)… and \(3x^2 + 2x - 15\ (= 0)\) not seen | Zero |
| One solution and \(3x^2 + 2x - 15\ (= 0)\) not seen | Zero |
| Missing brackets must be recovered | |
| \(\dfrac{3x^2 + 2x - 15}{(4x + 1)(x^2 + 3)} = 0\) followed by \(3x^2 + 2x - 15 = (4x + 1)(x^2 + 3)\) | M1M1M0M0A0 |
| \(\dfrac{3x^2 + 2x - 15}{(4x + 1)(x^2 + 3)} = 0\) followed by 1.93 and \(-2.59\) | M1M1M1M1A1 |