The diagram shows the floor of a new changing room. A square store cupboard fills one corner. The rest is used for changing. All lengths are in metres. stor...
Question 1 Report
The diagram shows the floor of a new changing room. A square store cupboard fills one corner. The rest is used for changing. All lengths are in metres.
- Expand and simplify \((3x - 2)(x + 1)\). (2)
- Expand \((x - 1)^2\). (1)
- Show that the changing floor is \((2x^2 + 3x - 3)\) m2. (2)
- That floor covers 24 m2. Show that \(2x^2 + 3x - 27 = 0\). (1)
- Solve this equation by factorising and say which root fits the room. (2)
Answer Details
The figure shows the whole floor as a rectangle \((3x - 2)\) m by \((x + 1)\) m, with a square store cupboard of side \((x - 1)\) m filling the bottom-right corner. The changing area is what remains after the cupboard is taken out.
(a) Multiplying each term of the first bracket by each term of the second:
\[(3x - 2)(x + 1) = 3x^2 + 3x - 2x - 2 = 3x^2 + x - 2\][2]
(b) A square bracket means the bracket multiplied by itself, not each term squared:
\[(x - 1)^2 = (x - 1)(x - 1) = x^2 - x - x + 1 = x^2 - 2x + 1\][1]
Writing \(x^2 - 1\) here is the classic error; the middle term \(-2x\) comes from the two cross products.
(c) The changing floor is the whole rectangle minus the square cupboard. Keep the second expression bracketed so every term changes sign:
\[(3x^2 + x - 2) - (x^2 - 2x + 1) = 3x^2 + x - 2 - x^2 + 2x - 1 = 2x^2 + 3x - 3 \text{ m}^2 \text{ as required.}\][2]
(d) Setting the expression equal to the given area and collecting on one side:
\[2x^2 + 3x - 3 = 24 \implies 2x^2 + 3x - 27 = 0 \text{ as required.}\][1]
(e) Two numbers are needed that multiply to \(2 \times (-27) = -54\) and add to \(3\), namely \(9\) and \(-6\), which leads to
\[(x - 3)(2x + 9) = 0\]so \(x = 3\) or \(x = -4.5\). [1]
The negative root is rejected because it would make every side length negative. The root that fits the room is \(x = 3\). [1]
Checking: the whole floor is \(7\) m by \(4\) m, giving \(28\) m\(^2\); the cupboard is \(2\) m by \(2\) m, giving \(4\) m\(^2\); and \(28 - 4 = 24\) m\(^2\) as stated.
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