Higher November 2023 Unit 2 Q19
19. Use the quadratic formula to solve the equation \(\dfrac{x + 3}{x} + 5 = 7x\).
Give your answers as decimal numbers.
You must show all your working. [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 |
|---|---|---|
| \(x + 3 + 5x = 7x^2\) OR \(x + 3 = x(7x - 5)\) OR \(x + 3 = 7x^2 - 5x\) | M1 | Clearing the denominator of \(x\). May be seen in stages, e.g. first writing \(\tfrac{x + 3}{x}\) as \(1 + \tfrac{3}{x}\), etc |
| \(7x^2 - 6x - 3 = 0\) | A1 | CAO '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 7, b = -6, \; c = -3\) used in the quadratic formula. |
| \(x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times (7) \times (-3)}}{2 \times (7)}\) | M1 | This substitution into the formula must be seen for M1, otherwise award M0A0A0. FT ‘their derived quadratic equation equated to zero’ provided of equivalent difficulty (\(a\), \(b\) and \(c\) must be non-zero). Allow one slip in substitution for M1 only, but must be correct formula. This can be awarded as a single attempt which may be seen anywhere in the solution for solving their quadratic equation equated to zero. |
| \(= \dfrac{3 \pm \sqrt{30}}{7} \left(\dfrac{6 \pm \sqrt{120}}{14}\right)\) | A1 | Can be implied from at least one correct value of \(x\) evaluated, provided M1 awarded. |
| \(x\) = 1·21(…) AND –0·35(…) | A1 | CAO for their quadratic equation. Both solutions required. Accept 1·2 and –0·4. |