Solutions — The Quadratic Formula

1. Identify a, b, c and calculate the discriminant. Fluency

   • (a) x² + 5x + 6 = 0: a = 1, b = 5, c = 6. Δ = 5² − 4(1)(6) = 25 − 24 = 1. (Perfect square ⇒ rational solutions.)
   • (b) 2x² − 3x − 5 = 0: a = 2, b = −3, c = −5. Δ = (−3)² − 4(2)(−5) = 9 + 40 = 49. (Perfect square ⇒ rational solutions.)
   • (c) x² − 4x + 4 = 0: a = 1, b = −4, c = 4. Δ = (−4)² − 4(1)(4) = 16 − 16 = 0. (One repeated solution.)
   • (d) 3x² + 2x + 1 = 0: a = 3, b = 2, c = 1. Δ = 2² − 4(3)(1) = 4 − 12 = −8. (Negative ⇒ no real solutions.)

2. Apply the quadratic formula — rational answers. Fluency

   • (a) x² − 5x + 6 = 0: a=1, b=−5, c=6. Δ = 25 − 24 = 1. x = (5 ± 1)/2. x = 3 or x = 2. (Note: this also factorises as (x−3)(x−2)=0 — the formula confirms it.)
   • (b) x² + x − 12 = 0: a=1, b=1, c=−12. Δ = 1 + 48 = 49. x = (−1 ± 7)/2. x = 6/2 = 3 or x = −8/2 = −4. x = 3 or x = −4
   • (c) 2x² − 7x + 3 = 0: a=2, b=−7, c=3. Δ = 49 − 24 = 25. x = (7 ± 5)/4. x = 12/4 = 3 or x = 2/4 = 1/2. x = 3 or x = 1/2
   • (d) 3x² + 5x − 2 = 0: a=3, b=5, c=−2. Δ = 25 + 24 = 49. x = (−5 ± 7)/6. x = 2/6 = 1/3 or x = −12/6 = −2. x = 1/3 or x = −2

3. Apply the formula — exact surd answers. Fluency

   • (a) x² − 4x + 1 = 0: a=1, b=−4, c=1. Δ = 16 − 4 = 12. √12 = 2√3. x = (4 ± 2√3)/2 = 2 ± √3
   • (b) x² + 6x + 3 = 0: a=1, b=6, c=3. Δ = 36 − 12 = 24. √24 = 2√6. x = (−6 ± 2√6)/2 = −3 ± √6
   • (c) 2x² − 2x − 1 = 0: a=2, b=−2, c=−1. Δ = 4 + 8 = 12. √12 = 2√3. x = (2 ± 2√3)/4 = (1 ± √3)/2. x = (1 + √3)/2 or x = (1 − √3)/2
   • (d) x² − 8x + 3 = 0: a=1, b=−8, c=3. Δ = 64 − 12 = 52. √52 = 2√13. x = (8 ± 2√13)/2 = 4 ± √13

4. Discriminant analysis — number and type of solutions. Fluency

   • (a) x² − 6x + 9 = 0: Δ = 36 − 36 = 0. One repeated rational solution. (x = 3)
   • (b) x² − 4x + 5 = 0: Δ = 16 − 20 = −4. No real solutions. (The parabola sits entirely above the x-axis.)
   • (c) x² − 6x + 7 = 0: Δ = 36 − 28 = 8. Δ > 0 but 8 is not a perfect square. Two distinct irrational (surd) solutions.
   • (d) 2x² − 7x + 3 = 0: Δ = 49 − 24 = 25. Δ > 0 and 25 is a perfect square. Two distinct rational solutions.

5. Choose method, then solve. Understanding

   • (a) x² − 9x + 20 = 0: Δ = 81 − 80 = 1 (perfect square). Factorise. Need two numbers that multiply to 20 and add to −9: use −4 and −5. (x − 4)(x − 5) = 0. x = 4 or x = 5
   • (b) x² − 3x − 1 = 0: Δ = 9 + 4 = 13 (not a perfect square). Quadratic formula. x = (3 ± √13)/2. x = (3 + √13)/2 or x = (3 − √13)/2 (≈ 3.30 or ≈ −0.30)
   • (c) 2x² + 5x − 12 = 0: Δ = 25 + 96 = 121 (perfect square: 11²). Factorise. ac = −24, need sum 5: use 8 and −3. 2x² + 8x − 3x − 12 = 0 ⇒ 2x(x + 4) − 3(x + 4) = 0 ⇒ (2x − 3)(x + 4) = 0. x = 3/2 or x = −4
   • (d) x² − 2x − 7 = 0: Δ = 4 + 28 = 32 (not a perfect square). Quadratic formula. √32 = 4√2. x = (2 ± 4√2)/2 = 1 ± 2√2

6. Projectile motion. Understanding

   • (a) Initial height: Substitute t = 0: h = −5(0)² + 20(0) + 25 = 25 m.
   • (b) When is h = 40 m?: Set h = 40: −5t² + 20t + 25 = 40. Rearrange: −5t² + 20t − 15 = 0. Divide by −5: t² − 4t + 3 = 0. Factorise: (t − 1)(t − 3) = 0. t = 1 s and t = 3 s. The ball passes through 40 m on the way up (t=1) and again on the way down (t=3).
   • (c) When does the ball land?: Set h = 0: −5t² + 20t + 25 = 0. Divide by −5: t² − 4t − 5 = 0. Factorise: (t − 5)(t + 1) = 0. t = 5 or t = −1. Reject t = −1 (time cannot be negative). The ball hits the ground at t = 5 seconds.

7. Diagonal of a rectangle. Understanding

   • (a) Quadratic equation: By Pythagoras: x² + (x + 3)² = 9². Expand: x² + x² + 6x + 9 = 81. Simplify: 2x² + 6x − 72 = 0, or equivalently x² + 3x − 36 = 0 (divide by 2).
   • (b) Solve by formula (exact form): Using x² + 3x − 36 = 0: a=1, b=3, c=−36. Δ = 9 + 144 = 153 = 9 × 17. √153 = 3√17. x = (−3 ± 3√17)/2. Reject the negative root (width cannot be negative). x = (−3 + 3√17)/2
   • (c) Dimensions to 2 d.p.: √17 ≈ 4.1231. x ≈ (−3 + 12.369)/2 ≈ 9.369/2 ≈ 4.68 cm. Length = x + 3 ≈ 7.68 cm. Check: 4.68² + 7.68² ≈ 21.9 + 59.0 = 80.9 ≈ 81 = 9² ✓

8. Discriminant with a parameter. Understanding

   • (a) Discriminant in terms of k: a=1, b=−k, c=k+3. Δ = (−k)² − 4(1)(k+3) = k² − 4k − 12. Factorise: Δ = (k − 6)(k + 2)
   • (b) Exactly one solution (k values): Δ = 0 ⇒ (k − 6)(k + 2) = 0. k = 6 or k = −2
   • (c) Two distinct real solutions (k values): Δ > 0 ⇒ (k − 6)(k + 2) > 0. This product is positive when both factors are positive (k > 6) or both are negative (k < −2). k < −2 or k > 6
   • (d) Solve when k = 8: Equation: x² − 8x + (8+3) = 0 ⇒ x² − 8x + 11 = 0. Δ = 64 − 44 = 20. √20 = 2√5. x = (8 ± 2√5)/2 = 4 ± √5

9. Rectangular fencing. Problem Solving

   • (a) Setting up the equation: Width = w. The two widths use 2w metres of fencing, leaving 80 − 2w for the length. Area = w(80 − 2w) = 750. Expand: 80w − 2w² = 750. Rearrange: 2w² − 80w + 750 = 0, or w² − 40w + 375 = 0 (divide by 2).
   • (b) Solve by formula: Using w² − 40w + 375 = 0: a=1, b=−40, c=375. Δ = 1600 − 1500 = 100. √100 = 10. w = (40 ± 10)/2. w = 25 m or w = 15 m (both exact; no decimal rounding needed).
   • (c) Both sets of dimensions: When w = 25 m: length = 80 − 2(25) = 30 m. Area = 25 × 30 = 750 m² ✓ When w = 15 m: length = 80 − 2(15) = 50 m. Area = 15 × 50 = 750 m² ✓ Two valid solutions: 25 m × 30 m or 15 m × 50 m.

10. A number and its reciprocal. Problem Solving

    • (a) Standard form equation: x + 1/x = 29/10. Multiply both sides by 10x: 10x² + 10 = 29x. Rearrange: 10x² − 29x + 10 = 0
    • (b) Apply the formula: a=10, b=−29, c=10. Δ = (−29)² − 4(10)(10) = 841 − 400 = 441 = 21². x = (29 ± 21)/20. x = 50/20 = 5/2 or x = 8/20 = 2/5. x = 5/2 or x = 2/5. Both are rational (discriminant was a perfect square).
    • (c) Verify both solutions: x = 5/2: 5/2 + 2/5 = 25/10 + 4/10 = 29/10 ✓   x = 2/5: 2/5 + 5/2 = 4/10 + 25/10 = 29/10 ✓   Both solutions are valid. Note: they are reciprocals of each other — if x is a solution, so is 1/x.