Solving Quadratic Equations — Solutions

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1. Solve by factorising. Fluency

   1. (a) x² − 7x + 12 = 0 (x − 3)(x − 4) = 0 → x = 3 or x = 4.
   2. (b) x² + x − 6 = 0 (x + 3)(x − 2) = 0 → x = −3 or x = 2.
   3. (c) 2x² + 5x − 3 = 0 (2x − 1)(x + 3) = 0 → x = 1/2 or x = −3.
   4. (d) 3x² − 12x = 0 3x(x − 4) = 0 → x = 0 or x = 4.
   5. (e) x² − 9 = 0 (x − 3)(x + 3) = 0 → x = 3 or x = −3.
   6. (f) 4x² − 4x + 1 = 0 (2x − 1)² = 0 → x = 1/2 (repeated root).

2. Solve by completing the square. Fluency

   1. (a) x² + 4x − 3 = 0 (x + 2)² − 4 − 3 = 0 → (x + 2)² = 7 → x = −2 ± √7.
   2. (b) x² − 6x + 1 = 0 (x − 3)² − 9 + 1 = 0 → (x − 3)² = 8 → x = 3 ± 2√2.
   3. (c) x² + 2x − 5 = 0 (x + 1)² − 1 − 5 = 0 → (x + 1)² = 6 → x = −1 ± √6.
   4. (d) x² − 8x + 10 = 0 (x − 4)² − 16 + 10 = 0 → (x − 4)² = 6 → x = 4 ± √6.

3. Quadratic formula. Fluency

   1. (a) x² − 4x + 1 = 0 Δ = 16 − 4 = 12. x = (4 ± 2√3)/2 = 2 ± √3.
   2. (b) 2x² + 3x − 2 = 0 Δ = 9 + 16 = 25. x = (−3 ± 5)/4. So x = 1/2 or x = −2.
   3. (c) 3x² − 5x + 1 = 0 Δ = 25 − 12 = 13. x = (5 ± √13)/6.
   4. (d) x² + x + 3 = 0 Δ = 1 − 12 = −11 < 0. No real solutions — the parabola has no x-intercepts and lies entirely above the x-axis.

4. Discriminant. Fluency

   1. (a) x² − 6x + 9 = 0 Δ = 36 − 36 = 0. One repeated real solution (perfect square).
   2. (b) 2x² + x + 4 = 0 Δ = 1 − 32 = −31 < 0. No real solutions.
   3. (c) x² − 3x − 5 = 0 Δ = 9 + 20 = 29 > 0. Two distinct real (irrational) solutions.
   4. (d) 4x² − 12x + 9 = 0 Δ = 144 − 144 = 0. One repeated real solution.

5. Rearrange and solve. Fluency

   1. (a) x(x + 5) = 14 x² + 5x − 14 = 0 → (x + 7)(x − 2) = 0 → x = −7 or x = 2.
   2. (b) (x + 1)(x − 3) = 5 x² − 2x − 3 = 5 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2.
   3. (c) x² = 3x + 10 x² − 3x − 10 = 0 → (x − 5)(x + 2) = 0 → x = 5 or x = −2.
   4. (d) 2x(x − 1) = x + 3 2x² − 2x − x − 3 = 0 → 2x² − 3x − 3 = 0.
      Δ = 9 + 24 = 33. x = (3 ± √33)/4.

6. Exact surd solutions. Understanding

   1. (a) x² − 2x − 4 = 0 Δ = 4 + 16 = 20. x = (2 ± √20)/2 = (2 ± 2√5)/2 = 1 ± √5.
   2. (b) 2x² + 2x − 3 = 0 Δ = 4 + 24 = 28. x = (−2 ± √28)/4 = (−2 ± 2√7)/4 = (−1 ± √7)/2.
      Written as: x = (−1 + √7)/2 or x = (−1 − √7)/2. (Denominators already rational.)
   3. (c) Derive general form via completing the square. x² + bx + c = 0
      x² + bx = −c
      (x + b/2)² − b²/4 = −c
      (x + b/2)² = b²/4 − c
      x + b/2 = ±√(b²/4 − c)
      x = −b/2 ± √(b²/4 − c) ✓

7. Equations with a parameter. Understanding

   1. (a) x² + mx + 9 = 0, exactly one solution. Δ = m² − 36 = 0 → m² = 36 → m = 6 or m = −6.
   2. (b) x² − 4x + k = 0, two distinct real solutions. Δ = 16 − 4k > 0 → k < 4.
   3. (c) kx² + 2x + k = 0, repeated root. Δ = 4 − 4k² = 0 → k² = 1 → k = 1 or k = −1.
      Note: k ≠ 0 (otherwise not a quadratic).

8. Quadratics in geometry. Understanding

   1. (a) Rectangle, perimeter 24 cm, area 32 cm². Let length = x. Width = 12 − x (since 2(x + w) = 24 → x + w = 12).
      Area: x(12 − x) = 32 → 12x − x² = 32 → x² − 12x + 32 = 0.
      (x − 4)(x − 8) = 0 → x = 4 or x = 8.
      Dimensions: 8 cm × 4 cm.
   2. (b) Right triangle, hypotenuse 10 cm, one leg 2 cm longer than the other. Let shorter leg = x. Longer leg = x + 2.
      Pythagoras: x² + (x + 2)² = 100.
      x² + x² + 4x + 4 = 100 → 2x² + 4x − 96 = 0 → x² + 2x − 48 = 0.
      Δ = 4 + 192 = 196. x = (−2 ± 14)/2 → x = 6 (taking positive value).
      Legs: 6 cm and 8 cm. Check: 6² + 8² = 36 + 64 = 100 = 10² ✓

9. Vieta’s formulas. Problem Solving

   1. (a) Prove α + β = −b, αβ = c. If α and β are roots: (x − α)(x − β) = x² − (α+β)x + αβ.
      Compare with x² + bx + c: coefficient of x gives −(α+β) = b → α+β = −b; constant gives αβ = c. ✓
   2. (b) Sum and product of roots of 3x² − 5x − 2 = 0. α + β = −(−5)/3 = 5/3. αβ = −2/3.
   3. (c) α + β = 5, αβ = 3; find p, q, and the roots. x² − px + q = 0, so p = α+β = 5, q = αβ = 3.
      Equation: x² − 5x + 3 = 0. Roots: x = (5 ± √13)/2.
   4. (d) One root is twice the other: β = 2α. Sum: α + 2α = 3α = 7 → α = 7/3, β = 14/3.
      Product: α × 2α = 2α² = k → k = 2(49/9) = 98/9.
      Roots: 7/3 and 14/3.

10. Equations reducible to quadratics. Problem Solving

    1. (a) x&sup4; − 5x² + 4 = 0 Let u = x²: u² − 5u + 4 = 0 → (u−1)(u−4) = 0 → u = 1 or u = 4.
       x² = 1 → x = ±1; x² = 4 → x = ±2.
       Solutions: x = −2, −1, 1, 2.
    2. (b) x&sup4; − 13x² + 36 = 0 Let u = x²: u² − 13u + 36 = 0 → (u−4)(u−9) = 0 → u = 4 or u = 9.
       x = ±2 or x = ±3.
    3. (c) (x² − 2x)² − 11(x² − 2x) + 24 = 0 Let u = x² − 2x: u² − 11u + 24 = 0 → (u−3)(u−8) = 0 → u = 3 or u = 8.
       Case 1: x² − 2x = 3 → x² − 2x − 3 = 0 → (x−3)(x+1) = 0 → x = 3 or x = −1.
       Case 2: x² − 2x = 8 → x² − 2x − 8 = 0 → (x−4)(x+2) = 0 → x = 4 or x = −2.
       Solutions: x = −2, −1, 3, 4.
    4. (d) x + 4/x = 5 Multiply by x (x ≠ 0): x² + 4 = 5x → x² − 5x + 4 = 0 → (x−1)(x−4) = 0.
       x = 1 or x = 4.