Higher June 2022 Paper 1 Q19
19 Solve \(\dfrac{1}{2x - 1} + \dfrac{3}{x - 1} = 1\)
Give your answer in the form \(\dfrac{p \pm \sqrt{q}}{2}\) where \(p\) and \(q\) are integers. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(\dfrac{5 \pm \sqrt{15}}{2}\) | M1 | for using a common denominator eg \(\dfrac{x - 1}{(2x - 1)(x - 1)} + \dfrac{3(2x - 1)}{(2x - 1)(x - 1)}\ (= 1)\) or \((x - 1) + 3(2x - 1) = (2x - 1)(x - 1)\) |
| M1 | for expanding and rearranging to get \(2x^2 - 10x + 5\ (= 0)\) | |
| M1 | (dep M1) ft for a method to solve their 3 term quadratic equation eg \(\dfrac{10 \pm \sqrt{(-10)^2 - 4 \times 2 \times 5}}{2 \times 2}\) or \(\dfrac{10 \pm \sqrt{60}}{4}\) or \(2\left[\left(x - \dfrac{5}{2}\right)^2 - \left(\dfrac{5}{2}\right)^2\right] + 5 = 0\) oe | |
| A1 | cao |
Additional guidance
Note we don’t need to see “= 0”; just the LHS is sufficient
Accept other forms of the 3 term quadratic, eg \(2x^2 - 10x = -5\)
Correct use of formula or completing the square