Higher June 2022 Paper 3 Q19
19 Here is the plan of the floor of an L-shaped room.
All lengths are in metres.
Not drawn accurately
(a) The area of the floor is 75 m2
Show that \(\quad x^2 + x - 90 = 0\) [3 marks]
(b) By factorising \(\quad x^2 + x - 90 \quad\) work out the value of \(x\).
You must show your working [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 – horizontal split | ||
| \(x(x - 2)\) and \(3(x - 5)\) | M1 | oe may be seen as two areas |
| \(x^2 - 2x + 3x - 15\) (= 75) | M1dep | oe expression with all brackets expanded |
| \(x^2 - 2x + 3x - 15 = 75\) and \(x^2 + x - 90 = 0\) or \(x^2 + x - 15 = 75\) and \(x^2 + x - 90 = 0\) | A1 | with full working seen |
| Alternative method 2 – vertical split | ||
| \((x - 5)(x + 1)\) and \(5(x - 2)\) | M1 | oe may be seen as two areas |
| \(x^2 - 5x + x - 5 + 5x - 10\) (= 75) or \(x^2 - 4x - 5 + 5x - 10\) (= 75) | M1dep | oe expression with all brackets expanded |
| \(x^2 - 5x + x - 5 + 5x - 10 = 75\) and \(x^2 + x - 90 = 0\) or \(x^2 - 4x - 5 + 5x - 10 = 75\) and \(x^2 + x - 90 = 0\) | A1 | with full working seen |
| Alternative method 3 – large rectangle subtract \(3 \times 5\) | ||
| \(x(x + 1)\) and \(3 \times 5\) | M1 | oe may be seen as two areas |
| \(x^2 + x - 15\) (= 75) | M1dep | oe expression with brackets expanded and \(3 \times 5\) evaluated |
| \(x^2 + x - 15 = 75\) and \(x^2 + x - 90 = 0\) | A1 | with full working seen |
| Alternative method 4 – split into three areas | ||
| \(3(x - 5)\) and \((x - 2)(x - 5)\) and \(5(x - 2)\) | M1 | oe may be seen as three areas |
| \(3x - 15 + x^2 - 2x - 5x + 10 + 5x - 10\) (= 75) or \(3x - 15 + x^2 - 7x + 10 + 5x - 10\) (= 75) | M1dep | oe expression with all brackets expanded |
| \(3x - 15 + x^2 - 2x - 5x + 10 + 5x - 10 = 75\) and \(x^2 + x - 90 = 0\) or \(3x - 15 + x^2 - 7x + 10 + 5x - 10 = 75\) and \(x^2 + x - 90 = 0\) | A1 | with full working seen |
Additional guidance
Ignore attempts to solve the equation or substituting values for \(x\)
Condone missing end bracket for M1
Condone missing pairs of brackets if recovered
eg \(3 \times x - 5\) recovered to \(3x - 15\)
| Answer | Mark | Comments |
|---|---|---|
| \((x - 9)(x + 10)\) (= 0) and answer 9 | B2 | B1 \((x - 9)(x + 10)\) (= 0) and answer 9 and \(-10\) SC1 \((x + 9)(x - 10)\) (= 0) and answer 10 |
Additional guidance
| If no response is seen, check part (a) for any creditworthy work | |
| Answer 9 with no working can be awarded up to B2 from correct factorising seen in part (a) | |
| Answer 9 from quadratic formula or completing the square | B1 |
| Answer 9 and \(-10\) from quadratic formula or completing the square | B0 |
| Answer from trial and improvement only | B0 |