Higher November 2020 Unit 2 Q19
19. Use the quadratic formula to solve \((5x + 3)(5x - 3) = 19x\).
Give your answers correct to 2 decimal places.
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 |
|---|---|---|
| Sight of \(25x^2 + 15x - 15x - 9\) | B1 | Or equivalent. |
| \(25x^2 - 19x - 9 = 0\) | B1 | '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 25, b = -19\), \(c = -9\) used in the quadratic formula. |
| \(x = \dfrac{-(-19) \pm \sqrt{(-19)^2 - 4 \times 25 \times (-9)}}{2 \times 25}\) | M1 | This substitution into the formula must be seen for M1, otherwise award M0A0A0. FT ‘their derived quadratic equation’ of equivalent difficulty (\(a\), \(b\) and \(c\) must be non-zero). Allow one slip in substitution for M1 only, but must be correct formula. |
| \(x = \dfrac{19 \pm \sqrt{1261}}{50}\) | A1 | Can be implied from at least one correct value of \(x\) evaluated, provided M1 awarded. |
| \(x = 1{\cdot}09\) with \(x = -0{\cdot}33\) (answers to 2dp) | A1 | CAO for their quadratic equation. |