Simplifying Surds — Solutions

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1. Surd or not a surd? Fluency

   1. (a) √16 Not a surd. √16 = 4, which is rational.
   2. (b) √7 Surd. 7 has no perfect square factor (other than 1), so √7 is irrational.
   3. (c) √0.25 Not a surd. √0.25 = 0.5, which is rational.
   4. (d) √11 Surd. 11 is prime so √11 is irrational.
   5. (e) √(4/9) Not a surd. √(4/9) = 2/3, which is rational.
   6. (f) √50 Surd (in its original form, √50 is irrational). Note: it simplifies to 5√2, but remains a surd.

2. Simplify each surd. Fluency

   1. (a) √12 √12 = √(4 × 3) = 2√3
   2. (b) √45 √45 = √(9 × 5) = 3√5
   3. (c) √75 √75 = √(25 × 3) = 5√3
   4. (d) √98 √98 = √(49 × 2) = 7√2
   5. (e) √200 √200 = √(100 × 2) = 10√2
   6. (f) √147 √147 = √(49 × 3) = 7√3

3. Collect like surds. Fluency

   1. (a) 3√5 + 7√5 10√5
   2. (b) 9√3 − 4√3 5√3
   3. (c) 2√7 + 5√7 − √7 6√7
   4. (d) 4√11 − 6√11 + 3√11 √11
   5. (e) 5√2 + √2 − 3√2 3√2

4. Simplify surds then collect like terms. Fluency

   1. (a) √8 + √18 = 2√2 + 3√2 = 5√2
   2. (b) √12 + √27 = 2√3 + 3√3 = 5√3
   3. (c) √50 − √8 = 5√2 − 2√2 = 3√2
   4. (d) 2√45 + √20 = 2(3√5) + 2√5 = 6√5 + 2√5 = 8√5
   5. (e) √75 − √48 + √3 = 5√3 − 4√3 + √3 = 2√3

5. Simplify each product. Fluency

   1. (a) √5 × √5 5
   2. (b) √3 × √12 √36 = 6
   3. (c) 2√7 × 3√7 6 × 7 = 42
   4. (d) 4√3 × 2√5 8√15
   5. (e) 3√2 × √8 3√16 = 3 × 4 = 12
   6. (f) 5√6 × 2√6 10 × 6 = 60

6. Mixed simplification. Understanding

   1. (a) √50 − 2√8 + √18 = 5√2 − 2(2√2) + 3√2 = 5√2 − 4√2 + 3√2 = 4√2
   2. (b) 3√12 + √75 − 2√27 = 3(2√3) + 5√3 − 2(3√3) = 6√3 + 5√3 − 6√3 = 5√3
   3. (c) √98 + √50 − √72 = 7√2 + 5√2 − 6√2 = 6√2
   4. (d) 2√80 − √45 + 3√20 = 2(4√5) − 3√5 + 3(2√5) = 8√5 − 3√5 + 6√5 = 11√5

7. Expanding with surds. Understanding

   1. (a) √3(2 + √3) = 2√3 + √3 × √3 = 2√3 + 3
   2. (b) √2(3√2 − √8) = 3√4 − √16 = 3(2) − 4 = 6 − 4 = 2
   3. (c) 2√5(3 + √5) = 6√5 + 2√5 × √5 = 6√5 + 2(5) = 6√5 + 10
   4. (d) √6(2√6 − √3 + 1) = 2√36 − √18 + √6 = 12 − 3√2 + √6

8. Perimeter and area. Understanding

   1. (a) Simplify √20. √20 = √(4 × 5) = 2√5 cm
   2. (b) Perimeter. Length = 3√5, Width = 2√5.
      Perimeter = 2(3√5 + 2√5) = 2(5√5) = 10√5 cm
   3. (c) Area. Area = 3√5 × 2√5 = 6 × 5 = 30 cm²

9. Diagonal of a rectangle. Problem Solving

   1. (a) Simplify the dimensions. √48 = 4√3 m; √75 = 5√3 m.
   2. (b) Diagonal by Pythagoras. d² = (√48)² + (√75)² = 48 + 75 = 123.
      d = √123 = √(3 × 41) = √3 × √41 m.
      (Cannot simplify further since 41 is prime.)
   3. (c) Perimeter as k√3. Perimeter = 2(4√3 + 5√3) = 2(9√3) = 18√3 m.
      k = 18. Both dimensions are integer multiples of √3, so their sum is also a multiple of √3.

10. Proof and surds. Problem Solving

    1. (a) Expand (√a + √b)². (√a + √b)² = (√a)² + 2√a√b + (√b)² = a + 2√(ab) + b
    2. (b) Exact value of (√3 + √12)². a = 3, b = 12: (√3 + √12)² = 3 + 2√36 + 12 = 15 + 2(6) = 15 + 12 = 27.
    3. (c) Show (√5 + √20)² = 45. (√5 + √20)² = 5 + 2√100 + 20 = 25 + 20 = 45 ✓
       This tells us √5 + √20 = √45 = 3√5. Indeed: √5 + 2√5 = 3√5. The identity confirms the simplification.