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Quadratic Functions and Equations in One Variable · Form 4

Quadratic Functions and Equations in One Variable: Worked Examples (KBAT)

This set hides quadratics inside real situations, a garden, a kicked ball and a photo border, so you must set up the equation yourself before solving. Best for students ready to turn words into maths.

Written by the spmmath.com.my editorial team.

Worked example 1

A rectangular vegetable garden is 3 m longer than it is wide. Its area is 40 m².

Find the width and the length of the garden.

  1. Let the width be x m; then the length is (x + 3) m.
  2. Area = width × length, so x(x + 3) = 40.
  3. Expand and rearrange: x² + 3x − 40 = 0.
  4. Factorise: (x + 8)(x − 5) = 0, so x = −8 or x = 5.
  5. A width cannot be negative, so reject x = −8; the width is x = 5 m.
  6. The length is x + 3 = 5 + 3 = 8 m.

Worked example 2

A ball is kicked straight up. Its height h metres after t seconds is h = 20t − 5t².

(a) After how many seconds does the ball return to the ground? (b) What is the maximum height reached?

  1. (a) At the ground h = 0, so 20t − 5t² = 0.
  2. Factorise: 5t(4 − t) = 0, so t = 0 or t = 4.
  3. t = 0 is the moment of the kick; the ball returns to the ground at t = 4 s.
  4. (b) Complete the square: h = −5t² + 20t = −5(t² − 4t) = −5((t − 2)² − 4).
  5. Expand: h = −5(t − 2)² + 20, so the maximum value is 20 when t = 2.
  6. The greatest height is 20 m, reached after 2 seconds.

Worked example 3

A rectangular photo measures 20 cm by 15 cm. It is placed in a mount with a border of uniform width x cm all around.

The total area of the photo and border together is 500 cm². Find the width of the border.

  1. The border adds x on each side, so the outer rectangle is (20 + 2x) by (15 + 2x) cm.
  2. Total area: (20 + 2x)(15 + 2x) = 500.
  3. Expand: 300 + 40x + 30x + 4x² = 500, so 4x² + 70x + 300 = 500.
  4. Rearrange: 4x² + 70x − 200 = 0, then divide by 2: 2x² + 35x − 100 = 0.
  5. Factorise: (2x − 5)(x + 20) = 0, so x = 2.5 or x = −20.
  6. A width cannot be negative, so reject x = −20; the border width is x = 2.5 cm.

Worked example 4

A right-angled triangle has a base that is 7 cm longer than its height. Its area is 30 cm².

Find the height and the base.

  1. Let the height be h cm; then the base is (h + 7) cm.
  2. Area of a triangle = ½ × base × height, so ½ × (h + 7) × h = 30.
  3. Multiply both sides by 2: h(h + 7) = 60, i.e. h² + 7h − 60 = 0.
  4. Factorise: (h + 12)(h − 5) = 0, so h = −12 or h = 5.
  5. A length cannot be negative, so h = 5; base = 5 + 7 = 12.

Worked example 5

The product of two consecutive positive even numbers is 168. Find the two numbers.

  1. Let the smaller even number be n; the next even number is n + 2.
  2. Their product: n(n + 2) = 168, i.e. n² + 2n − 168 = 0.
  3. Factorise: (n + 14)(n − 12) = 0, so n = −14 or n = 12.
  4. The numbers are positive, so n = 12 and n + 2 = 14.

Worked example 6

A shopkeeper spends RM 60 buying identical notebooks. If each notebook had cost RM 1 less, he could have bought 5 more for the same RM 60.

Find the original price of one notebook.

  1. Let the original price be RM x per notebook, so the number bought is 60/x.
  2. At RM (x − 1) each he could buy 60/(x − 1), which is 5 more: 60/(x − 1) = 60/x + 5.
  3. Multiply through by x(x − 1): 60x = 60(x − 1) + 5x(x − 1).
  4. Simplify: 60x = 60x − 60 + 5x² − 5x, so 5x² − 5x − 60 = 0, i.e. x² − x − 12 = 0.
  5. Factorise: (x − 4)(x + 3) = 0, so x = 4 or x = −3; a price is positive, so x = 4.

Worked example 7

A durian orchard is planted in equal rows. Each row has 5 more trees than the number of rows.

Altogether there are 176 trees in the orchard. Find the number of rows and the number of trees in each row.

  1. Let the number of rows be r. Then the number of trees per row is r + 5.
  2. Total trees: r(r + 5) = 176, so r² + 5r − 176 = 0
  3. Solve: r = [−5 ± √(25 + 704)] / 2 = [−5 ± √729] / 2 = [−5 ± 27] / 2
  4. Reject the negative root; r = 11 (the number of rows must be positive)
  5. Trees per row = 11 + 5 = 16; check: 11 × 16 = 176 ✓

Worked example 8

A farmer wants to fence a rectangular plot next to a straight river, so no fencing is needed along the riverside. Using 80 m of fencing for the other three sides, the farmer wants to enclose an area of 750 m².

Find the possible dimensions of the plot.

  1. Let the width perpendicular to the river be x m. The two widths use 2x m of fencing, leaving (80 − 2x) m for the side parallel to the river.
  2. Area: x(80 − 2x) = 750, so 80x − 2x² = 750, giving 2x² − 80x + 750 = 0, or x² − 40x + 375 = 0
  3. Solve: x = [40 ± √(1600 − 1500)] / 2 = [40 ± √100] / 2 = [40 ± 10] / 2
  4. x = 25 or x = 15
  5. When x = 15: length = 80 − 30 = 50, giving 15 m × 50 m. When x = 25: length = 80 − 50 = 30, giving 25 m × 30 m. Check: 15 × 50 = 750 ✓ and 25 × 30 = 750 ✓

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

How do I turn a word problem into a quadratic equation?

Identify the unknown, define it clearly as x, then translate the relationships in the problem (area, perimeter, product, and so on) into an equation. KBAT questions reward a correctly formed equation even before solving, write your variable definition and equation-forming steps clearly.

What do examiners look for in optimisation questions using quadratics?

They want you to express the quantity to be maximised or minimised as a quadratic function, then use completing the square to find the turning point. Marks come from correctly identifying whether the turning point gives a maximum or minimum value in the context of the problem.

Why do KBAT quadratic questions often reject one of the two roots?

Because real-world quantities like length, time, or price cannot be negative. After solving, always check both roots against the problem's context and reject any value that doesn't make physical sense, examiners specifically test whether you remember this reasoning step.

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