Surds and Indices in SSC CGL: What You Must Know
Surds and indices questions in SSC CGL and CHSL test three skills: using the laws of indices, simplifying or rationalising surds, and comparing roots of different orders. They are short, scoring questions once the rules below are automatic.
Key takeaway: aᵐ × aⁿ = aᵐ⁺ⁿ; aᵐ ÷ aⁿ = aᵐ⁻ⁿ; (aᵐ)ⁿ = aᵐⁿ; a⁰ = 1; a⁻ⁿ = 1/aⁿ; a^(p/q) = q-th root of aᵖ. To rationalise 1/(√a − √b), multiply by (√a + √b). To compare ∛4, √2 and ⁶√6, raise all to the 6th power (LCM of 3, 2, 6).
Every answer below was computed with Python (sympy). Revise identities in our algebra formulas for SSC CGL alongside this note.
Laws of Indices (Exponents) Table
| Law | Rule | Example |
|---|---|---|
| Product | aᵐ × aⁿ = aᵐ⁺ⁿ | 2³ × 2⁴ = 2⁷ = 128 |
| Quotient | aᵐ ÷ aⁿ = aᵐ⁻ⁿ | 5⁶ ÷ 5⁴ = 25 |
| Power of a power | (aᵐ)ⁿ = aᵐⁿ | (3²)³ = 3⁶ = 729 |
| Power of a product | (ab)ⁿ = aⁿbⁿ | (2 × 5)³ = 1000 |
| Zero index | a⁰ = 1 (a ≠ 0) | 7⁰ = 1 |
| Negative index | a⁻ⁿ = 1/aⁿ | 2⁻³ = 1/8 |
| Fractional index | a^(p/q) = (q-th root of a)ᵖ | 27^(2/3) = 3² = 9 |
| Equal bases | aˣ = aʸ ⇒ x = y | 2ˣ = 32 ⇒ x = 5 |
Surds: Rules and Simplification
A surd is an irrational root such as √2, ∛5. Rules:
- √a × √b = √(ab); √a ÷ √b = √(a/b).
- Like surds add: 3√2 + 5√2 = 8√2. Unlike surds (√2 + √3) cannot be combined.
- Conjugate: (√a + √b)(√a − √b) = a − b. Use it to rationalise.
- (√a + √b)² = a + b + 2√(ab). Example: (√3 + √2)² = 5 + 2√6.
- Square root of a + 2√b: find two numbers with sum a and product b. √(7 + 4√3) = √(7 + 2√12) = 2 + √3 (4 + 3 = 7, 4 × 3 = 12).
How to Compare Surds of Different Orders
Raise every surd to the LCM of the root orders, then compare whole numbers.
Example: which is greatest, ∛4, √2 or ⁶√6? LCM of 3, 2, 6 = 6. Sixth powers: (∛4)⁶ = 16, (√2)⁶ = 8, (⁶√6)⁶ = 6. So ∛4 is greatest and ⁶√6 is smallest.
Solved Surds and Indices Questions (SSC Pattern)
- 27^(2/3) = 9.
- 16^(−3/4) = 1/(2³) = 1/8.
- (1/32)^(−2/5) = 32^(2/5) = 2² = 4.
- (0.04)^(−1.5) = (1/25)^(−3/2) = 25^(3/2) = 125.
- 2¹⁰ × 2⁻⁷ ÷ 2² = 2¹ = 2.
- If 3^(x + 1) = 81, then x = 3.
- If 5^(x − 2) = 125, then x = 5.
- Rationalise 1/(√5 − √3): multiply by (√5 + √3): (√5 + √3)/2.
- If x = 3 + 2√2, then 1/x = 3 − 2√2, so x + 1/x = 6 and x² + 1/x² = 6² − 2 = 34.
- √(12 + √(12 + √(12 + …))) = y gives y² = 12 + y, so y = 4 (positive root).
- √(7 + 4√3) = 2 + √3.
- Greatest of ∛4, √2, ⁶√6: ∛4.
Tricks for Surds and Indices
- Infinite nested roots: √(a + √(a + …)) = the positive root of y² − y − a = 0. If a = n(n + 1), the answer is n + 1 (12 = 3 × 4, so 4).
- √(a√(a√a…)) infinitely = a.
- x = a + √b with a² − b = 1 means 1/x = a − √b, so x + 1/x = 2a.
- Same exponent, compare bases; same base, compare exponents (for base > 1).
Common Mistakes
- Writing √(a + b) = √a + √b. It is false (√9 + √16 = 7, but √25 = 5).
- Forgetting that a negative index means a reciprocal, not a negative number.
- Comparing surds of different orders without raising to a common power.
- Taking the negative root in nested-root questions (lengths and roots here are positive).
Surds and Indices FAQs
What are the laws of indices?
aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ and a^(p/q) is the q-th root of aᵖ.
How do you rationalise a surd?
Multiply the numerator and denominator by the conjugate. For 1/(√a − √b), multiply by (√a + √b) to get (√a + √b)/(a − b).
How do you compare surds like ∛4 and √2?
Raise both to the LCM of the orders (here 6): 16 and 8, so ∛4 is greater.
Are surds and indices asked in SSC CGL?
Yes, usually as short simplification questions in Tier 1 and in Tier 2 algebra and number-system sets.
Practice Next
- Topic test: SSC CGL Surds and Indices Mock Test 03.
- Mock tests: SSC CGL Quantitative Aptitude Sectional Mock Test 05, SSC CGL Algebra Mock Test 01 and SSC CHSL Quantitative Aptitude Sectional Mock Test 04.
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- Exam details: SSC CGL 2026.
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