Solutions — Solving Quadratic Equations by Factorising

1. Each quadratic equation is already in standard form. Solve by factorising. Fluency

   • (a) x² + 5x + 6 = 0: Need two numbers that multiply to 6 and add to 5: use 2 and 3. (x + 2)(x + 3) = 0. x = −2 or x = −3
   • (b) x² − 7x + 12 = 0: Need two numbers that multiply to 12 and add to −7: use −3 and −4. (x − 3)(x − 4) = 0. x = 3 or x = 4
   • (c) x² − 4x − 21 = 0: Need two numbers that multiply to −21 and add to −4: use −7 and 3. (x − 7)(x + 3) = 0. x = 7 or x = −3
   • (d) x² − 16 = 0: Difference of two squares: (x − 4)(x + 4) = 0. x = 4 or x = −4
   • (e) x² − 6x + 9 = 0: Perfect square: (x − 3)² = 0. One repeated root. x = 3
   • (f) x² + 3x = 0: HCF is x: x(x + 3) = 0. x = 0 or x = −3

2. Rearrange each equation to standard form (= 0), then solve by factorising. Fluency

   • (a) x² = 4x + 12: Rearrange: x² − 4x − 12 = 0. Need −6 and 2. (x − 6)(x + 2) = 0. x = 6 or x = −2
   • (b) x² + 10 = 7x: Rearrange: x² − 7x + 10 = 0. Need −5 and −2. (x − 2)(x − 5) = 0. x = 2 or x = 5
   • (c) 2x² = 8x: Rearrange: 2x² − 8x = 0. HCF is 2x: 2x(x − 4) = 0. x = 0 or x = 4
   • (d) 3x² + 5x = 2: Rearrange: 3x² + 5x − 2 = 0. Product-sum: ac = −6, need sum 5; use 6 and −1. 3x² + 6x − x − 2 = 0 ⇒ 3x(x + 2) − 1(x + 2) = 0 ⇒ (3x − 1)(x + 2) = 0. x = 1/3 or x = −2

3. Solve each non-monic quadratic equation. Fluency

   • (a) 2x² + 7x + 3 = 0: ac = 6, need sum 7: use 6 and 1. 2x² + 6x + x + 3 = 0 ⇒ 2x(x + 3) + 1(x + 3) = 0 ⇒ (2x + 1)(x + 3) = 0. x = −1/2 or x = −3
   • (b) 3x² − 5x − 2 = 0: ac = −6, need sum −5: use −6 and 1. 3x² − 6x + x − 2 = 0 ⇒ 3x(x − 2) + 1(x − 2) = 0 ⇒ (3x + 1)(x − 2) = 0. x = −1/3 or x = 2
   • (c) 4x² − 9 = 0: Difference of two squares: (2x − 3)(2x + 3) = 0. x = 3/2 or x = −3/2
   • (d) 6x² − x − 2 = 0: ac = −12, need sum −1: use −4 and 3. 6x² − 4x + 3x − 2 = 0 ⇒ 2x(3x − 2) + 1(3x − 2) = 0 ⇒ (2x + 1)(3x − 2) = 0. x = −1/2 or x = 2/3

4. Take out the HCF first, then solve. Fluency

   • (a) 3x² − 12x = 0: HCF = 3x: 3x(x − 4) = 0. x = 0 or x = 4
   • (b) 5x² − 45 = 0: HCF = 5: 5(x² − 9) = 0 ⇒ 5(x − 3)(x + 3) = 0. x = 3 or x = −3
   • (c) 2x² + 6x + 4 = 0: HCF = 2: 2(x² + 3x + 2) = 0 ⇒ 2(x + 1)(x + 2) = 0. x = −1 or x = −2
   • (d) 4x² − 16x + 16 = 0: HCF = 4: 4(x² − 4x + 4) = 0 ⇒ 4(x − 2)² = 0. Repeated root. x = 2

5. Find and fix the student’s error. Understanding

   • (a) Identify the error: The student divided both sides by x. This is invalid because x might equal 0 — dividing by 0 is undefined. When you divide by x you assume x ≠ 0, which discards a valid solution. You must never divide by a variable to “cancel” it.
   • (b) Correct solution: Rearrange: x² − 5x = 0. Factorise using HCF: x(x − 5) = 0. Apply Null Factor Law: x = 0 or x − 5 = 0. x = 0 or x = 5
   • (c) Lost solution: x = 0 was lost. Both x = 0 and x = 5 satisfy the original equation: 0² − 5(0) = 0 ✓ and 25 − 25 = 0 ✓.

6. Positive dimensions. Understanding

   • (a) Quadratic equation: By Pythagoras: x² + (x − 3)² = (x + 3)². Expand: x² + x² − 6x + 9 = x² + 6x + 9. Simplify: x² − 12x = 0.
   • (b) Solve for x: x(x − 12) = 0, so x = 0 or x = 12. Since x is a length it must be positive, so reject x = 0. x = 12
   • (c) Side lengths: Legs: x = 12 cm and x − 3 = 9 cm. Hypotenuse: x + 3 = 15 cm. Check: 12² + 9² = 144 + 81 = 225 = 15² ✓ (This is a 3–4–5 triangle scaled by 3.)

7. Consecutive integers. Understanding

   • (a) Equation for the product: The two consecutive integers are n and (n + 1). Their product: n(n + 1) = 56
   • (b) Rearrange and solve: n² + n − 56 = 0. Need two numbers that multiply to −56 and add to 1: use 8 and −7. (n + 8)(n − 7) = 0. n = −8 or n = 7. Since the integers are positive, n = 7. n = 7
   • (c) The two integers: 7 and 8. Check: 7 × 8 = 56 ✓

8. Rectangular area from a quadratic. Understanding

   • (a) Quadratic equation: Width = w, length = w + 4. Area = w(w + 4) = 96. w² + 4w − 96 = 0
   • (b) Solve for the width: Need two numbers that multiply to −96 and add to 4: use 12 and −8. (w + 12)(w − 8) = 0. w = −12 or w = 8. Reject w = −12 (width cannot be negative). Width = 8 m
   • (c) Dimensions and verification: Width = 8 m, Length = 8 + 4 = 12 m. Check: 8 × 12 = 96 m² ✓

9. Solve each equation, rearranging and expanding as needed. Problem Solving

   • (a) (x + 3)(x − 1) = 5: Expand: x² + 2x − 3 = 5. Rearrange: x² + 2x − 8 = 0. Factorise: (x + 4)(x − 2) = 0. x = −4 or x = 2
   • (b) (2x − 1)(x + 4) = (x + 2)(x + 3): Expand LHS: 2x² + 7x − 4. Expand RHS: x² + 5x + 6. Rearrange: x² + 2x − 10 = 0. Check discriminant: b² − 4ac = 4 + 40 = 44 — not a perfect square, so this does not factorise over the integers. Preview of L03: the quadratic formula gives x = (−2 ± √44) / 2 ≈ 2.32 or −4.32. The skills to solve this neatly come in the next lesson.
   • (c) x(x + 5) = (x + 1)(x + 2) + 2: Expand LHS: x² + 5x. Expand RHS: x² + 3x + 2 + 2 = x² + 3x + 4. Set equal: x² + 5x = x² + 3x + 4. The x² terms cancel: 2x = 4. x = 2 (the equation reduces to a linear equation!)

10. Projectile motion. Problem Solving

    • (a) When does the ball hit the ground?: Set h = 0: −5t² + 20t + 60 = 0. Divide by −5: t² − 4t − 12 = 0. Factorise: (t − 6)(t + 2) = 0. t = 6 or t = −2
    • (b) Which solution to reject?: Reject t = −2 because time cannot be negative. The ball hits the ground at t = 6 seconds.
    • (c) Heights at t = 1 and t = 3: h(1) = −5(1)² + 20(1) + 60 = −5 + 20 + 60 = 75 m. h(3) = −5(9) + 20(3) + 60 = −45 + 60 + 60 = 75 m. Both give 75 m — the parabola is symmetric about t = 2 (the vertex), so t = 1 and t = 3 are equidistant from the vertex and reach the same height.