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Chapter 3 Practice Paper: Quadratic Expressions and Equations — mark scheme

QuizQuipPractice Series
QQ/S2MATH/CH3

Secondary 2 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Quadratic Expressions and Equations

Chapter 3 Practice Paper · Maximum mark: 30

Notes for markers

  • Award one mark for each point marked [1], up to the marks shown for the part.
  • Accept answers worded differently that show the same understanding.
  • Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

QuestionAnswer
1CThe partial products x², 2x, 6x and 12 collect to x² + 8x + 12.1
2AThe square identity keeps the middle term: x² − 2·3·x + 9.1
3DA difference of two squares splits into one plus bracket and one minus bracket.1
4B−5 and 3 multiply to −15 and add to −2.1
5CBoth terms share 3x; taking out the full common factor leaves x − 4.1
6AEach bracket takes its turn at zero: x − 2 = 0 or x + 7 = 0.1
7DSquaring either 10 or −10 gives 100, so both are solutions.1
8BOn the x-axis y = 0, so the brackets give x = −1 or x = 9.1
9C3x² + 12x − 2x − 8 collects to 3x² + 10x − 8.1
10Ah = 0 needs t = 0 or 20 − t = 0, so the ball starts and lands at those times.1

Section B: Structured questions[20 marks]

QuestionAnswer and marking pointsMarks
1(a)

x², −9x, 4x and −36.

  • All four products seen [1]
  • Signs carried correctly [1]
2
1(b)

x² − 5x − 36 [1]

1
1(c)

Left: (6)(−7) = −42. Right: 4 − 10 − 36 = −42 ✓.

  • Left side evaluates to −42 [1]
  • Right side evaluates to −42 and matches [1]
2
Total for Question 15
2(a)

3x(2x − 5) [1]

1
2(b)

(x + 3)(x + 8), since 3 and 8 multiply to 24 and add to 11.

  • Factor pair 3 and 8 identified [1]
  • (x + 3)(x + 8) [1]
2
2(c)

(x + 9)(x − 9); the difference of two squares.

  • (x + 9)(x − 9) [1]
  • Difference of two squares named [1]
2
Total for Question 25
3(a)

(x + 4)(x − 3), since 4 and −3 multiply to −12 and add to 1.

  • Factor pair 4 and −3 identified [1]
  • (x + 4)(x − 3) [1]
2
3(b)

x = −4 or x = 3.

  • x = −4 [1]
  • x = 3 [1]
2
3(c)

For x = 3: 9 + 3 − 12 = 0 ✓.

  • Substitution reaches 0 [1]
1
Total for Question 35
4(a)

(x + 6)(x − 2) = 20 expands to x² + 4x − 12 = 20, so x² + 4x − 32 = 0.

  • Expands (x + 6)(x − 2) = 20 [1]
  • Rearranges to x² + 4x − 32 = 0 [1]
2
4(b)

(x + 8)(x − 4) = 0, so x = −8 or x = 4.

  • (x + 8)(x − 4) = 0 [1]
  • x = −8 or 4 [1]
2
4(c)

x = 4 (the other root makes the width negative), giving 10 cm by 2 cm.

  • Chooses x = 4 with dimensions 10 cm by 2 cm [1]
1
Total for Question 45

— End of mark scheme —