Higher November 2017 Unit 2 Q9
9.
(a) Show that \((10w + 3)(w - 1) - (2 - 3w)^2 \equiv w^2 + 5w - 7\). [4]
(b) Use the quadratic formula to solve the equation \(w^2 + 5w - 7 = 0\).
Give your answers correct to 2 decimal places. [3]
Give your answers correct to 2 decimal places. [3]
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 |
|---|---|---|
| \(10w^2 - 10w + 3w - 3\) | B1 | Or equivalent. |
| \(4 - 6w - 6w + 9w^2\) | B1 | Or equivalent. |
| \(10w^2 - 10w + 3w - 3 - 4 + 6w + 6w - 9w^2\) | B1 | FT if at least B1 awarded for equivalent level of difficulty, ie. at least three terms for each expansion. Penalise any further error. |
| \((=) w^2 + 5w - 7\) | B1 | CAO (convincing). Dependent on B1B1B1. |
| Scheme | Marks | Comments |
|---|---|---|
| Trial and improvement method gains M0. | ||
| \(w = \dfrac{-(5) \pm \sqrt{(5)^2 - 4 \times 1 \times (-7)}}{2 \times 1}\) | M1 | Allow one slip in substitution, but must be correct formula. |
| \(= \dfrac{-5 \pm \sqrt{53}}{2}\) | A1 | |
| \(w = 1.14\) AND \(w = -6.14\) | A1 | CAO |