Hint
Area is length times width. Test both algebraic roots against \(x>0\).
Worked solution
(a)
Multiply width by length: \(x(2x+3)=35\). Expanding and subtracting 35 gives the required equation.
Answer: \(x(2x+3)=35\), hence \(2x^2+3x-35=0\).
(b)
\((2x-7)(x+5)=0\), giving \(\displaystyle x= \frac{7}{2}\) or \(x=-5\). Reject \(-5\) because \(x>0\). Then length \(\displaystyle = 2(\frac{7}{2})+ 3= 10\) m.
Answer: Width \(3.5\text{ m}\); length \(10\text{ m}\).
(c)
\(P=2(3.5+10)=27\) m.
Answer: \(27\text{ m}\).
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official examination-board mark scheme. Award stated criteria once each, up to the part maximum. Accept valid equivalent methods. Follow-through applies only where explicitly stated; do not award an accuracy criterion for an incorrect final result.
(a) · 2 marks
- 1 mark: Form the correct area equation.
- 1 mark: Expand and rearrange correctly.
Alternative methods: An annotated expression for length times width is sufficient.
Common error: Using perimeter instead of area.
(b) · 3 marks
- 1 mark: Correctly solve the quadratic, obtaining both algebraic roots.
- 1 mark: Reject the negative root and state width 3.5 m.
- 1 mark: Find length 10 m.
Alternative methods: Quadratic formula followed by the domain check is accepted.
Common error: Reporting both roots as dimensions.
(c) · 2 marks
- 1 mark: Use twice the sum of the two dimensions.
- 1 mark: Obtain 27 m; follow through positive dimensions from part (b) for method credit.
Alternative methods: Adding all four side lengths is accepted.
Common error: Giving square metres as the perimeter unit.
Total: 7 marks.