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Algebraic Identities & Elementary Surds
Quantitative Aptitude · Algebra, Equations & Graphs
Useful for: SSC CGL
Overview
Combining identities with surd rules simplifies expressions like (√a + √b)² or 1/(√a − √b) in one step.
Key points
- (√a + √b)(√a − √b) = a − b.
- (√a + √b)² = a + b + 2√(ab).
- Rationalising: multiply numerator and denominator by the conjugate.
- If x = a + √b, then 1/x can be found by rationalising.
- Comparing surds: square both sides when both are positive.
Explanation
- x = √3 + 1 → x² = 4 + 2√3.
- For x + 1/x with x = 2 + √3: 1/x = 2 − √3, so x + 1/x = 4.
- √(a + 2√b) can be written as √m + √n when m + n = a and mn = b.
- To compare √3 + √5 and √2 + √6, square both: 8 + 2√15 vs 8 + 2√12.
Step by step
- Spot the conjugate or identity.
- Rationalise or expand.
- Simplify and approximate to check.
Examples
- (√5 + √3)(√5 − √3) = 5 − 3 = 2.
- 6/(√5 − √2) = 6(√5 + √2)/(5 − 2) = 2(√5 + √2).
- x = 2 + √3 → x + 1/x = (2 + √3) + (2 − √3) = 4 → x² + 1/x² = 16 − 2 = 14.
Traps and advanced points
- Trap: Writing √(a + b) as √a + √b.
- Trap: Rationalising with the same sign instead of the conjugate.
- Trap: Dropping the 2√(ab) term when squaring.
Quick revision
- Conjugate removes the root.
- x = 2 + √3 → x + 1/x = 4.
- Square both sides to compare surds.
Practice MCQs
If a + b = 5 and ab = 6, then a² + b² is
- 13
- 25
- 19
- 11
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Answer: (a) 13. (a + b)² − 2ab = 25 − 12.
If x + 1/x = 3, then x² + 1/x² is
- 7
- 9
- 11
- 5
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Answer: (a) 7. (x + 1/x)² − 2 = 9 − 2.
The value of (√3 + √2)(√3 − √2) is
- 1
- 5
- √6
- 0
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Answer: (a) 1. (a + b)(a − b) = a² − b² = 3 − 2.
Rationalising 1/(√5 − 2) gives
- √5 − 2
- √5 + 2
- (√5 + 2)/9
- 1/3
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Answer: (b) √5 + 2. Multiply by √5 + 2; the denominator becomes 5 − 4 = 1.
If a + b = 4 and ab = 3, then a³ + b³ is
- 28
- 64
- 36
- 52
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Answer: (a) 28. (a + b)³ − 3ab(a + b) = 64 − 36.
√48 in simplest form is
- 6√2
- 3√4
- 2√12
- 4√3
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Answer: (d) 4√3. 48 = 16 × 3.
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