NCEA Level 2Indices and surdsAS 91261

Indices and surds: 7 exam questions solved step by step

What this lesson covers

  • Index laws apply to every factor inside a bracket, so a power of a product raises each part.
  • A surd can be written as a fractional index, which lets the index laws do the work.
Key idea

7 questions, one toolkit

  • 1. A power hits every factor in a bracket.
  • 2. Roots are fractional indices.
  • 3. Take square factors out of surds.
  • 4. Same base: compare the indices.
Your turn

Question 1 · Achieved

  • Simplify (5x²y)³
Worked example

Q1: (5x²y)³

  • Cube every factor.
  • Including the 5.
  1. (5x²y)³ = 5³ × (x²)³ × y³every factor
  2. 5³ = 5 × 5 × 5 = 125
  3. (x²)³ = x⁶multiply the indices
  4. = 125x⁶y³
  5. 5x⁶y³the 5 was not cubed
Your turn

Question 2 · Achieved

  • Simplify √75 + √48 − √27
Worked example

Q2: √75 + √48 − √27

  • Largest square factor.
  • Then collect like surds.
  1. √75 = √25 × √3 = 5√3square factor 25
  2. √48 = √16 × √3 = 4√3square factor 16
  3. √48 = √4 × √12 = 2√12√12 still simplifies
  4. √27 = √9 × √3 = 3√3square factor 9
  5. √75 + √48 − √27
  6. = 5√3 + 4√3 − 3√3substitute
  7. = (5 + 4 − 3)√3like surds
  8. = 6√3
Your turn

Question 3 · Merit

  • Simplify (2a³b⁻²)³ ÷ (4a⁻¹b)
  • Give the answer with positive indices.
Worked example

Q3: (2a³b⁻²)³ ÷ (4a⁻¹b)

  • Bracket first.
  • Divide: subtract indices.
  1. (2a³b⁻²)³ = 2³a⁹b⁻⁶every factor cubed
  2. = 8a⁹b⁻⁶
  3. 8a⁹b⁻⁶ ÷ 4a⁻¹b
  4. 8 ÷ 4 = 2
  5. a: 9 − (−1) = 10subtract the indices
  6. b: −6 − 1 = −7b means b¹
  7. = 2a¹⁰b⁻⁷
  8. = 2a¹⁰/b⁷negative: one over
Your turn

Question 4 · Merit

  • Write x²√x ÷ ∛x
  • as a single power of x.
Worked example

Q4: x²√x ÷ ∛x

  • Roots become powers.
  • Common denominator: 6.
  1. √x = x^(1/2), ∛x = x^(1/3)roots as indices
  2. x² × x^(1/2) ÷ x^(1/3)
  3. = x^(2 + 1/2 − 1/3)add, then subtract
  4. 2 + 1/2 − 1/3 = 12/6 + 3/6 − 2/6
  5. = 13/6
  6. x²√x ÷ ∛x = x^(13/6)
  7. x = 64: 4096 × 8 ÷ 4 = 8192
  8. 64^(13/6) = 2¹³ = 8192
Your turn

Question 5 · Merit, in context

  • A school garden bed is a rectangle
  • with area 30 m² and width √6 m.
  • Find its length and its perimeter,
  • in simplest surd form.
Worked example

Q5: length and perimeter

  • No surd underneath.
  • Answer with units.
  1. length = 30 ÷ √6 = 30/√6area ÷ width
  2. = 30/√6 × √6/√6rationalise
  3. = 30√6/6√6 × √6 = 6
  4. = 5√6 m30 ÷ 6 = 5
  5. 5√6 × √6 = 5 × 6 = 30
  6. P = 2(5√6 + √6)two lengths, two widths
  7. = 2 × 6√6 = 12√6 mlike surds
  8. Length 5√6 m, perimeter 12√6 m
Your turn

Question 6 · Excellence

  • Solve 4^x × 8^(x − 1) = 32
Worked example

Q6: 4^x × 8^(x − 1) = 32

  • Everything as a power of 2.
  • Same base: equal indices.
  1. 4 = 2², 8 = 2³, 32 = 2⁵common base 2
  2. (2²)^x × (2³)^(x − 1) = 2⁵
  3. 2^(2x) × 2^(3x − 3) = 2⁵power of a power
  4. 2^(2x + 3x − 3) = 2⁵multiply: add
  5. 2^(5x − 3) = 2⁵collect
  6. 5x − 3 = 5same base, equal
  7. 5x = 8, x = 8/5 = 1.6
  8. 4^1.6 × 8^0.6 = 32
Your turn

Question 7 · Excellence

  • Show that (3ⁿ⁺² − 3ⁿ) ÷ (4 × 3ⁿ⁺¹)
  • has the same value for every n,
  • and find that value.
Worked example

Q7: the same for every n

  • Split off 3ⁿ.
  • A common factor cancels.
  1. 3ⁿ⁺² = 3ⁿ × 3² = 9 × 3ⁿmultiply: add
  2. 3ⁿ⁺¹ = 3ⁿ × 3 = 3 × 3ⁿthe same way
  3. top: 9 × 3ⁿ − 3ⁿ = 8 × 3ⁿnine lots less one
  4. bottom: 4 × 3 × 3ⁿ = 12 × 3ⁿ
  5. (8 × 3ⁿ) ÷ (12 × 3ⁿ)
  6. = 8/12 = 2/33ⁿ cancels
  7. n = 1: (27 − 3) ÷ 36 = 24/36
  8. Always 2/3, whatever n is
Summary

The toolkit that solved all seven

  • Every factor, roots as indices, one base, then check.
StepHow
Bracketsevery factor gets the power
Roots√x = x^(1/2), ∛x = x^(1/3)
Dividesubtract; minus a minus adds
Surdstake out the largest square
Rationalise× √a/√a, top and bottom
Same basethen equate the indices

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