Rationalising Denominators — Solutions

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1. Rationalise each denominator and simplify. Fluency

   1. (a) 1/√3 Multiply top and bottom by √3: (1 × √3)/(√3 × √3) = √3/3
   2. (b) 5/√5 5/√5 × √5/√5 = 5√5/5 = √5
   3. (c) 2/√7 2/√7 × √7/√7 = 2√7/7
   4. (d) √3/√2 √3/√2 × √2/√2 = √6/2
   5. (e) 6/√6 6/√6 × √6/√6 = 6√6/6 = √6
   6. (f) 4/(3√2) 4/(3√2) × √2/√2 = 4√2/(3 × 2) = 4√2/6 = 2√2/3

2. State the conjugate of each expression. Fluency

   1. (a) 2 + √5 Conjugate: 2 − √5 (flip the sign between the terms)
   2. (b) √3 − 1 Conjugate: √3 + 1
   3. (c) √7 + √2 Conjugate: √7 − √2
   4. (d) 4 − √11 Conjugate: 4 + √11
   5. (e) 3√2 + 5 Conjugate: 3√2 − 5

3. Expand each product using the difference of two squares identity. Fluency

   1. (a) (2 + √3)(2 − √3) = 2² − (√3)² = 4 − 3 = 1
   2. (b) (√5 + 1)(√5 − 1) = (√5)² − 1² = 5 − 1 = 4
   3. (c) (√7 − √2)(√7 + √2) = (√7)² − (√2)² = 7 − 2 = 5
   4. (d) (3 + 2√5)(3 − 2√5) = 3² − (2√5)² = 9 − 4 × 5 = 9 − 20 = −11
   5. (e) (4√3 − 1)(4√3 + 1) = (4√3)² − 1² = 16 × 3 − 1 = 48 − 1 = 47

4. Rationalise each denominator and simplify. Fluency

   1. (a) 1/(1 + √2) Multiply by (1 − √2)/(1 − √2):
      Numerator: 1 × (1 − √2) = 1 − √2
      Denominator: 1² − (√2)² = 1 − 2 = −1
      Result: (1 − √2)/(−1) = √2 − 1
   2. (b) 3/(4 − √5) Multiply by (4 + √5)/(4 + √5):
      Numerator: 3(4 + √5) = 12 + 3√5
      Denominator: 4² − (√5)² = 16 − 5 = 11
      Result: (12 + 3√5)/11
   3. (c) √2/(√2 + 1) Multiply by (√2 − 1)/(√2 − 1):
      Numerator: √2(√2 − 1) = 2 − √2
      Denominator: (√2)² − 1² = 2 − 1 = 1
      Result: 2 − √2
   4. (d) 6/(√3 + √2) Multiply by (√3 − √2)/(√3 − √2):
      Numerator: 6(√3 − √2)
      Denominator: (√3)² − (√2)² = 3 − 2 = 1
      Result: 6(√3 − √2) = 6√3 − 6√2
   5. (e) (√5 + 2)/(√5 − 2) Multiply by (√5 + 2)/(√5 + 2):
      Numerator: (√5 + 2)² = 5 + 4√5 + 4 = 9 + 4√5
      Denominator: (√5)² − 2² = 5 − 4 = 1
      Result: 9 + 4√5

5. Rationalise each denominator and express in simplest form. Fluency

   1. (a) 10/√5 10/√5 × √5/√5 = 10√5/5 = 2√5
   2. (b) √18/(2√3) First simplify √18 = 3√2.
      So: 3√2/(2√3) × √3/√3 = 3√6/(2 × 3) = 3√6/6 = √6/2
   3. (c) (3 + √2)/(3 − √2) Multiply by (3 + √2)/(3 + √2):
      Numerator: (3 + √2)² = 9 + 6√2 + 2 = 11 + 6√2
      Denominator: 3² − (√2)² = 9 − 2 = 7
      Result: (11 + 6√2)/7
   4. (d) (√6 − √2)/√2 Multiply by √2/√2:
      Numerator: (√6 − √2) × √2 = √12 − √4 = 2√3 − 2
      Denominator: 2
      Result: (2√3 − 2)/2 = √3 − 1

6. Rationalising in geometry. Understanding

   1. (a) Rationalise 12/√3. 12/√3 × √3/√3 = 12√3/3 = 4√3 cm
   2. (b) Perimeter of the rectangle. Length = 4√3 cm, Width = √3 cm.
      Perimeter = 2(4√3 + √3) = 2 × 5√3 = 10√3 cm
   3. (c) Verification. Length × Width = 4√3 × √3 = 4 × 3 = 12 cm² ✓

7. Equivalent expressions. Understanding

   1. (a) Rationalise 2/(√6 − √2). Conjugate of (√6 − √2) is (√6 + √2).
      Multiply: 2(√6 + √2) / [(√6)² − (√2)²] = 2(√6 + √2) / (6 − 2) = 2(√6 + √2)/4
      = (√6 + √2)/2
   2. (b) Is this equal to Alicia's answer (√6 + √2)/2? Yes. Our answer is (√6 + √2)/2, which is exactly Alicia's answer. Alicia is correct.
   3. (c) Is Ben's answer (√3 + 1)/√2 equivalent? Rationalise Ben's expression: (√3 + 1)/√2 × √2/√2 = (√6 + √2)/2.
      This equals Alicia's and our answer. Ben's answer is also correct — just written in a different (un-rationalised) form.

8. Solving equations with surd denominators. Understanding

   1. (a) x = 1/√2 + √2. Rationalise 1/√2: 1/√2 × √2/√2 = √2/2.
      x = √2/2 + √2 = √2/2 + 2√2/2 = 3√2/2
   2. (b) 3/(2 + √3) = x − √3. Find x. Rationalise 3/(2 + √3): multiply by (2 − √3)/(2 − √3).
      Numerator: 3(2 − √3) = 6 − 3√3
      Denominator: 2² − (√3)² = 4 − 3 = 1
      So left side = 6 − 3√3.
      6 − 3√3 = x − √3
      x = 6 − 3√3 + √3 = 6 − 2√3

9. Show that the following expressions are equal. Problem Solving

   1. (a) Show that 1/(√n + √(n+1)) = √(n+1) − √n. Start with left side: 1/(√n + √(n+1)).
      Multiply by (√(n+1) − √n)/(√(n+1) − √n):
      Numerator: √(n+1) − √n
      Denominator: (√(n+1))² − (√n)² = (n+1) − n = 1
      Result: (√(n+1) − √n)/1 = √(n+1) − √n ✓
   2. (b) Evaluate the telescoping sum. Using part (a), each term 1/(√k + √(k+1)) = √(k+1) − √k.
      So the sum = (√2 − √1) + (√3 − √2) + (√4 − √3) + … + (√9 − √8).
      All intermediate terms cancel (telescoping): result = √9 − √1 = 3 − 1 = 2.

10. Resistance in electrical circuits. Problem Solving

    1. (a) Write 1/R⊂1; and rationalise 1/R⊂2;. 1/R⊂1; = 1/√3 = √3/3 (rationalised).
       1/R⊂2; = 1/(1 + √3) × (1 − √3)/(1 − √3) = (1 − √3)/(1 − 3) = (1 − √3)/(−2) = (√3 − 1)/2
    2. (b) Find 1/R = 1/R⊂1; + 1/R⊂2;. 1/R = √3/3 + (√3 − 1)/2.
       Common denominator 6: = 2√3/6 + 3(√3 − 1)/6 = (2√3 + 3√3 − 3)/6 = (5√3 − 3)/6
    3. (c) Find R by taking the reciprocal and rationalising. R = 6/(5√3 − 3).
       Rationalise: multiply by (5√3 + 3)/(5√3 + 3).
       Denominator: (5√3)² − 3² = 75 − 9 = 66.
       Numerator: 6(5√3 + 3) = 30√3 + 18.
       R = (30√3 + 18)/66 = (5√3 + 3)/11
    4. (d) R as a decimal to 3 significant figures. R = (5√3 + 3)/11 ≈ (5 × 1.732 + 3)/11 = (8.660 + 3)/11 = 11.660/11 ≈ 1.06 ohms (3 s.f.).
       This combined resistance is less than either individual resistor (R⊂1; ≈ 1.73Ω, R⊂2; ≈ 2.73Ω), which is expected — parallel resistors always give a lower total resistance.