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Algebraic Identities

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Useful for: SSC CGL

BOOK-GUIDED ORIGINAL SSC MATHS PRACTICE — Algebraic Identities

FOUNDATION

(a + b)² − (a − b)² = 4ab.

METHOD

Identify the quantities given, choose the exact relationship, calculate in the same unit, then check the answer against the question.

WORKED EXAMPLE 1

Calculate (34+21)² − (34−21)².

Using (a+b)²−(a−b)²=4ab, answer = 4×34×21=2856.

WORKED EXAMPLE 2

Calculate (13+24)² − (13−24)².

Using (a+b)²−(a−b)²=4ab, answer = 4×13×24=1248.

IMPORTANT EXCEPTION / COMMON TRAP

Remember the middle term in squaring a binomial.

QUICK REVISION

(a + b)² − (a − b)² = 4ab.

SEE MORE: https://study.kundali-iq.com/ssc-maths/identities/

For complete timed maths practice, use SSC CGL Test Series → New Maths 100 Mixed Sets or Chapter-wise sets. These are independent original questions, not copied previous-year questions.

Practice MCQs

Calculate (34+21)² − (34−21)².
1. 2856
2. 2666
3. 3046
4. 3236
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Answer: (a) 2856. Using (a+b)²−(a−b)²=4ab, answer = 4×34×21=2856.
Calculate (13+24)² − (13−24)².
1. 1414
2. 1248
3. 1331
4. 1165
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Answer: (b) 1248. Using (a+b)²−(a−b)²=4ab, answer = 4×13×24=1248.
Calculate (14+2)² − (14−2)².
1. 119
2. 112
3. 105
4. 126
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Answer: (b) 112. Using (a+b)²−(a−b)²=4ab, answer = 4×14×2=112.
Calculate (4+4)² − (4−4)².
1. 68
2. 60
3. 64
4. 72
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Answer: (c) 64. Using (a+b)²−(a−b)²=4ab, answer = 4×4×4=64.
If a = 7 and b = 3, find (a + b)² − (a − b)²
1. 16
2. 40
3. 100
4. 84
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Answer: (d) 84. (a + b)² − (a − b)² = 4ab = 4 × 7 × 3 = 84.

Algebraic Identities

Practice MCQs

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