Higher June 2019 Unit 2 Q19
19. Use the quadratic formula to solve \((3x - 2)^2 = (x + 1)(x + 2)\).
Give your answers correct to 2 decimal places.
You must show all your working. [6]
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 \(9x^2 - 6x - 6x + 4\) | B1 | Or equivalent. |
| Sight of \(x^2 + x + 2x + 2\) | B1 | Or equivalent. |
| \(8x^2 - 15x + 2 = 0\) | B1 | FT expansions of equivalent level of difficulty provided B1 previously awarded. '= 0' required, but may be implied by an attempt to use the quadratic formula or if \(a = 8, b = -15\), \(c = 2\) used in the quadratic formula. |
| \(x = \dfrac{-(-15) \pm \sqrt{(-15)^2 - 4 \times 8 \times 2}}{2 \times 8}\) | M1 | This substitution into the formula must be seen for M1. FT ‘their derived quadratic equation’ equated to zero of equivalent difficulty (\(a\), \(b\) and \(c\) must be non-zero). Allow one slip in substitution for M1 only, but must be correct formula. |
| \(x = \dfrac{15 \pm \sqrt{161}}{16}\) | A1 | Can be implied from at least one correct value of \(x\) evaluated. |
| \(x = 1.73\) with \(x = 0.14\) (answers to 2dp) | A1 | CAO for their quadratic equation but not if complex roots. M0A0A0 if trial and improvement used or for unsupported answers. |