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Grade 11 · Equations & Inequalities · 10 min read

Word Problems with Quadratics (Grade 11)

Turning a sentence into an equation, solving it, and — the step that costs the most marks — rejecting the answer that does not make sense in the context.

Word problems test whether you can build the equation, not just solve one. The marks split roughly: half for setting it up, half for solving and interpreting.

1The four steps

  • Define your variable in words: 'Let x be the width in metres.'
  • Write every other quantity in terms of x.
  • Form the equation from the sentence that gives a total, product or area.
  • Solve, then reject any answer that makes no sense and answer the question that was asked.
💡

Write the 'Let x be …' line even if you think it is obvious. It is a mark, every single time.

2Consecutive numbers

Worked ExampleThe product of two consecutive positive integers is 132. Find them.
  1. 1
    Let the smaller integer be x. The next one is x + 1.
    Consecutive means one apart.
  2. 2
    x(x + 1) = 132.
    'Product' means multiply.
  3. 3
    x² + x − 132 = 0.
    Expand and move to standard form.
  4. 4
    (x + 12)(x − 11) = 0, so x = −12 or x = 11.
    Factorise: −12 × 11 = −132 and −12 + 11 = −1… check: (x+12)(x−11) = x² + x − 132. ✓
  5. 5
    Reject x = −12 because the integers must be positive.
    This rejection line carries a mark.
  6. 6
    The integers are 11 and 12.
    11 × 12 = 132. ✓

3Area problems

Worked ExampleA rectangle is 3 m longer than it is wide and its area is 40 m². Find its dimensions.
  1. 1
    Let the width be x metres. Then the length is (x + 3) metres.
    Everything in terms of one variable.
  2. 2
    x(x + 3) = 40.
    Area = length × width.
  3. 3
    x² + 3x − 40 = 0.
  4. 4
    (x + 8)(x − 5) = 0, so x = −8 or x = 5.
  5. 5
    Reject x = −8 — a width cannot be negative.
  6. 6
    Width 5 m, length 8 m.
    5 × 8 = 40. ✓
⚠️

Answering 'x = 5' is not the answer to 'find its dimensions'. Go back and answer the actual question, with units.

4Speed, distance and time

time = distancespeed
Worked ExampleA cyclist rides 60 km. If she had gone 5 km/h faster the trip would have taken 1 hour less. Find her speed.
  1. 1
    Let her speed be x km/h. Time taken = 60x.
  2. 2
    Faster speed is (x + 5), taking 60x + 5.
  3. 3
    60x60x + 5 = 1.
    The faster trip is 1 hour SHORTER, so subtract in this order.
  4. 4
    Multiply by x(x + 5): 60(x + 5) − 60x = x(x + 5).
    Clear the fractions.
  5. 5
    300 = x² + 5x → x² + 5x − 300 = 0.
  6. 6
    (x + 20)(x − 15) = 0, so x = 15 (reject −20).
    Speed cannot be negative.
  7. 7
    Her speed is 15 km/h.
    Check: 60÷15 = 4 h and 60÷20 = 3 h — exactly 1 hour less. ✓

Practice exercises

Work each one out, then click to reveal the answer.

  1. 1
    The product of two consecutive positive integers is 132. Determine the integers.
    Show answer ▾
    x² + x − 132 = 0 → (x+12)(x−11) = 0 → reject x = −12 → 11 and 12
  2. 2
    A rectangle is 3 m longer than it is wide and has an area of 40 m². Determine its dimensions.
    Show answer ▾
    x(x+3) = 40 → x = 5 (reject −8) → 5 m by 8 m
  3. 3
    The sum of a number and its square is 56. Determine the number if it is positive.
    Show answer ▾
    x + x² = 56 → (x+8)(x−7) = 0 → 7 (reject −8)
  4. 4
    A cyclist rides 60 km. Riding 5 km/h faster would take 1 hour less. Determine her speed.
    Show answer ▾
    60x60x+5 = 1 → x² + 5x − 300 = 0 → 15 km/h
  5. 5
    Two consecutive even numbers have a product of 168. Determine them.
    Show answer ▾
    x(x+2) = 168 → x² + 2x − 168 = 0 → (x+14)(x−12) = 0 → 12 and 14
  6. 6
    Why must you write down a line rejecting one of the two answers?
    Show answer ▾
    Because a length, speed or count cannot be negative — the rejection line is worth a mark.
  7. 7
    A rectangle's length is twice its width. Its area is 72 m². Determine the width.
    Show answer ▾
    2x² = 72 → x² = 36 → 6 m (reject −6)
  8. 8
    What is the first line you should write in any word problem?
    Show answer ▾
    A definition of the variable, e.g. 'Let x be the width in metres'.
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