Algebra:
Algebra: Quadratics,
Quadratic Rearranging
Formulae & Identities| Maths:187 | Class:12th
Formulae and Identities
Practice Questions and Solutions
Exam Questions
Answer all questions fully.
1 A field is rectangular in shape. The lengths are given in metres.
4𝑥𝑥 − 3
2𝑥𝑥 + 4
1 (a) Find an expression for the perimeter of the field.
[2 marks]
Answer
1 (b) Tick the expression, for the area of the field
[2 marks]
18𝑥𝑥 − 12
(2𝑥𝑥 + 4)(4𝑥𝑥 − 3)
8𝑥𝑥 2 + 10𝑥𝑥 − 12
6𝑥𝑥 2 − 14𝑥𝑥 − 12
2 A playground is rectangular in shape. An expression is given for the area of the
playground in 𝑚𝑚2 .
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑥𝑥 2 + 7𝑥𝑥 − 18
2 (a) Find an expression for the perimeter of the playground.
[3 marks]
Answer
2 (b) Given 𝑥𝑥 = 11 find the value of the area of the playground in 𝑚𝑚2 .
[2 marks]
Answer
3 (a) Make 𝑥𝑥 the subject of 4𝑥𝑥 + 12 = 𝑎𝑎𝑎𝑎 + 8𝑦𝑦
[3 marks]
Answer
3 (b) Hence or otherwise, given 𝑎𝑎 = 2 and 𝑦𝑦 = 10, find the value of 𝑥𝑥.
[2 marks]
Answer
Solutions |Algebra: Quadratic Formulae & Identities |Maths:187
Algebra: Quadratics, Rearranging
Formulae and Identities
Exam Questions
Q Answer Mark Comments
1(a) 2𝑥𝑥 + 4 + 4𝑥𝑥 − 3 + 2𝑥𝑥 + 4 + 4𝑥𝑥 − 3 M1 oe
12𝑥𝑥 + 2 A1 Allow 𝑃𝑃 = ⋯
B1 Chooses one of (2𝑥𝑥 + 4)(4𝑥𝑥 − 3) or
1(b) B2
8𝑥𝑥 2 + 10𝑥𝑥 − 12
Chooses both of (2𝑥𝑥 + 4)(4𝑥𝑥 − 3) and
OR
8𝑥𝑥 2 + 10𝑥𝑥 − 12 and no incorrect
statements. Chooses both of (2𝑥𝑥 + 4)(4𝑥𝑥 − 3) and
8𝑥𝑥 2 + 10𝑥𝑥 − 12 and one incorrect
statement.
2(a) (𝑥𝑥 + 9)(𝑥𝑥 − 2) M1 Factorises
𝑥𝑥 + 9 + 𝑥𝑥 − 2 + 𝑥𝑥 + 9 + 𝑥𝑥 − 2 M1
4𝑥𝑥 + 14 A1
(11 + 9) × (11 − 2)
2(b) M1 Substitutes into either original
OR
2 quadratic or factorised form
11 + 7 × 11 − 18
180 A1
3(a) 4𝑥𝑥 − 𝑎𝑎𝑎𝑎 = 8𝑦𝑦 − 12 M1 Isolates 𝑥𝑥 terms
OR 𝑎𝑎𝑎𝑎 − 4𝑥𝑥 = 12 − 8𝑦𝑦
𝑥𝑥 (4 − 𝑎𝑎) = 8𝑦𝑦 − 12 M1 Factorises 𝑥𝑥 terms
OR 𝑥𝑥 (𝑎𝑎 − 4) = 12 − 8𝑦𝑦
8𝑦𝑦 − 12 A1
𝑥𝑥 =
4 − 𝑎𝑎
OR
12 − 8𝑦𝑦
𝑥𝑥 =
𝑎𝑎 − 4
3(b) 8 × 10 − 12 M1 Substitutes into either original or
𝑥𝑥 =
4−2 rearranged form o.e.
34 A1 cao
