June 2019 Paper 2 Q18
18 Solve \(\quad 4(x - 5)^2 = k^2 \quad\) where \(k\) is a constant.
Give your answers in their simplest form in terms of \(k\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(x - 5 = \dfrac{k}{2}\) or \(x - 5 = -\dfrac{k}{2}\) or \(2(x - 5) = k\) or \(2x - 10 = k\) or \(2(x - 5) = -k\) or \(2x - 10 = -k\) | M1 | oe linear equation eg \(x - 5 = \sqrt{\dfrac{k^2}{4}}\) or \(x = \dfrac{k}{2} + 5\) or \(\sqrt{4}\,(x - 5) = \sqrt{k^2}\) |
| \(x - 5 = \dfrac{k}{2}\) and \(x - 5 = -\dfrac{k}{2}\) or \(2(x - 5) = k\) and \(2(x - 5) = -k\) or \(2x - 10 = k\) and \(2x - 10 = -k\) | A1 | oe eg \(x - 5 = \pm\dfrac{k}{2}\) square root(s) must be processed implied by final A1 |
| \(\dfrac{k}{2} + 5\) and \(-\dfrac{k}{2} + 5\) | A1 | oe simplest form eg \(\dfrac{10 + k}{2}\) and \(\dfrac{10 - k}{2}\) or \(\dfrac{k + 10}{2}\) and \(\dfrac{k - 10}{-2}\) or \(5 \pm 0.5k\) |
| Alternative method 2 | ||
| \(4x^2 - 40x + 100 - k^2\ (= 0)\) | M1 | expands and collects terms |
| \(\dfrac{-40 \pm \sqrt{(-40)^2 - 4 \times 4 \times (100 - k^2)}}{2 \times 4}\) | A1 | oe eg \(\dfrac{40 \pm \sqrt{16k^2}}{8}\) or \(\dfrac{40 \pm 4k}{8}\) implied by final A1 |
| \(\dfrac{k}{2} + 5\) and \(-\dfrac{k}{2} + 5\) | A1 | oe simplest form eg \(\dfrac{10 + k}{2}\) and \(\dfrac{10 - k}{2}\) or \(\dfrac{k + 10}{2}\) and \(\dfrac{k - 10}{-2}\) or \(5 \pm 0.5k\) |
Additional guidance
| Allow recovery of missing brackets |