Higher November 2016 Unit 2 Q16
16. Use the quadratic formula to solve \((3x - 1)^2 = x(2x + 3) + 7\).
Give your answers correct to 2 decimal places. [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 |
|---|---|---|
| \((3x - 1)^2 = 9x^2 - 3x - 3x + 1\) | B1 | |
| \(\ldots\ = 2x^2 + 3x + 7\) | B1 | |
| \(7x^2 - 9x - 6 = 0\) | B1 | CAO. '= 0' may be implied in further working. |
| \((x =)\ \dfrac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 7 \times (-6)}}{2 \times 7}\) | M1 | FT ‘their derived quadratic equation’ set to zero and of equivalent level of difficulty (\(a\), \(b\) and \(c\) are non-zero). Allow one slip in substitution, but must be correct formula. |
| \(= \dfrac{9 \pm \sqrt{249}}{14}\) | A1 | If one slip seen or a positive \(b\) used award A0. |
| \(x = 1.77\) AND \(x = -0.48\) (answers to 2dp) | A1 | CAO |