Higher November 2016 Unit 1 Q14
14. Aled has three concrete slabs.
Two of the slabs are square, with each side of length \(x\) metres.
The third slab is rectangular and measures 1 metre by \((x\,+1)\) metres.
The three concrete slabs cover an area of 7 m2.
(a) Show that \(2x^2 + x - 6 = 0\). [1]
(b) Solve the equation to find the length of each side of the square slabs.
You must justify any decisions that you make. [4]
You must justify any decisions that you make. [4]
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 |
|---|---|---|
| \(2x^2 + x + 1 = 7\) OR \(x^2 + x^2 + x + 1 = 7\) leading to \(2x^2 + x - 6 = 0\) | B1 | Must be seen. Accept 1(x +1) for x + 1. |
| Scheme | Marks | Comments |
|---|---|---|
| \((2x - 3)(x + 2)\) (=0) | B2 | B1 for \((2x \ldots\ 3)(x \ldots\ 2)\) |
| (Length of each side =) 1·5 (metres) | B1 | F.T. from ‘their two brackets’. (If both F.T. solutions are of the same sign, then both are required for this B1.) Ignore presence of x = –2. |
| Statement about ignoring x = – 2 as would be a negative length. | E1 | F.T provided one solution is positive and the other is negative. |
Using quadratic formula.
| Scheme | Marks | Comments |
|---|---|---|
| (\(x\) =) \(\dfrac{-1 \pm \sqrt{[1^2 - 4(2)(-6)]}}{2(2)}\) | M1 | Allow one error, in sign or substitution, but not in the formula. |
| = \(\dfrac{-1 \pm \sqrt{49}}{4}\) | A1 | |
| x = 1·5 (metres) [ignore x = -2] | A1 |
Using trial and improvement
| Scheme | Marks | Comments |
|---|---|---|
| Award B3 for a method leading to both solutions, namely x = 1·5 and x = -2, otherwise B0. |