Surds & Indices

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These questions look intimidating and are usually mechanical. The whole method is: express every term as a power of one common base, then add and subtract exponents. Almost all the difficulty is in spotting the shared base.

Laws of Indices

am × an = am+n  ·  am ÷ an = am−n

(am)n = amn  ·  a0 = 1  ·  a−n = 1/an

a1/n = n√a  ·  am/n = n√(am)

Powers Worth Knowing on Sight

Base 2Base 3Base 4
2&sup5; = 32, 2⁶ = 643³ = 27, 3⁴ = 814³ = 64 = 4³
2⁷ = 128, 2⁸ = 2563⁵ = 24316 = 4², 256 = 4⁴
2⁹ = 512, 2¹⁰ = 1024—4096 = 4⁶ = 16³
Find the common base first. 256, 64 and 16 are all powers of 4 and of 2. Choosing the base that the answer is expressed in — here 4, because the question ends in 4K — saves a conversion at the end.

Surd Simplification

Rationalising: multiply numerator and denominator by the conjugate. 1/(√a + √b) × (√a − √b)/(√a − √b) = (√a − √b)/(a − b).

Nested surds: if a + 2√b can be written as (√x + √y)² with x + y = a and xy = b, then √(a + 2√b) = √x + √y.

Worked Examples from Real Papers

TS Police Constable 2022 — Q48

If (256)2.5 × (16)4.5 ÷ (64)1.6 = 4K, find 5K + 4.

Convert everything to base 4: 256 = 4⁴, 16 = 4², 64 = 4³.
(4⁴)2.5 = 410  ·  (4²)4.5 = 49  ·  (4³)1.6 = 44.8
Combining exponents: 10 + 9 − 4.8 = 14.2, so K = 14.2.
5K + 4 = 71 + 4 = 75.
Note: a fractional K is fine here because the question multiplies it by 5, which clears the decimal. An answer that looks untidy mid-way is not necessarily wrong.

TS Police SI 2022 — Q48

Evaluate ( ∛(−2197) × ∛(−125) ÷ ∛(27/512) ) + ⅔

Cube roots of negatives are negative: ∛(−2197) = −13 and ∛(−125) = −5.
Their product is (−13)(−5) = 65.
∛(27/512) = 3/8, since 27 = 3³ and 512 = 8³.
65 ÷ 3/8 = 65 × 8/3 = 520/3.
Adding ⅔: 520/3 + 2/3 = 522/3 = 174.
Method: take the cube root of numerator and denominator separately — recognising 512 = 8³ is what makes this a two-line question.

TS Police SI 2022 — Q49

If x is an integer satisfying 9x2/3 + 9x−2/3 = 82, find √(12 + 2√x).

Substitute t = x2/3. Then 9t + 9/t = 82, so 9t² − 82t + 9 = 0.
Solving: t = (82 ± 80)/18, giving t = 9 or t = 1/9.
x2/3 = 9 → x = 27. (The other root gives x = 1/27, not an integer, so it is rejected.)
Now √(12 + 2√27) = √(12 + 6√3).
Look for (√x + √y)²: since 3 + 9 = 12 and 3 × 9 = 27, we get (√3 + 3)² = 12 + 6√3.
Answer: 3 + √3.
Method: the substitution t = x2/3 turns an intimidating equation into an ordinary quadratic. Any expression of the form a·u + a/u = k should trigger it.

Check which root the question wants. Quadratic substitutions produce two values, and the question usually restricts you — "x is an integer" in the example above discards one of them. Skipping that line is how a correct calculation still yields the wrong option.

Also see: Number System · Mensuration · Percentages

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