Exercise 11.4 — Algebraic Identities
Algebraic identities and their applications.
What is an Algebraic Identity?
Exercise 11.4 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus) introduces one of the most powerful tools in algebra — algebraic identities. These are equations that remain true for every possible value of the variables involved. They allow you to find products and evaluate numbers mentally, without writing out full multiplication.
The key difference between an identity and a regular equation is this: an equation like a² − 4 = 0 is true only for a = 2 or a = −2. But an identity like a(a + 2) ≡ a² + 2a is true for every value of a — substitute 1, 5, −7, or any number and it always holds. That is why we write the special symbol ≡ (instead of =) between the two sides of an identity.
The Four Standard Identities — Derived and Explained
All four identities in this exercise come from direct binomial multiplication — exactly the method you practised in Exercise 11.3. The derivation below shows why they work, so you understand them rather than just memorise them.
Why (a + b)² gives a² + 2ab + b² — step by step
Why (a − b)(a + b) gives a² − b² — step by step
How to Select the Right Identity — Decision Guide
Every question in Exercise 11.4 asks you to select a suitable identity before solving. Use this checklist to decide which identity fits:
| What you see | Identity to use | Key signal |
|---|---|---|
| Two identical brackets multiplied together, both with + sign | (a+b)² ≡ a²+2ab+b² | Both brackets are the same with a + inside |
| Two identical brackets multiplied together, both with − sign | (a−b)² ≡ a²−2ab+b² | Both brackets are the same with a − inside |
| Two brackets with the same terms but one + and one − | (a+b)(a−b) ≡ a²−b² | One bracket has +, the other has −, rest is the same |
| Two brackets sharing a common base expression, with different constant add-ons | (x+a)(x+b) ≡ x²+(a+b)x+ab | Same "x" part but different constants a and b |
Question 1 — Select a Suitable Identity and Find the Products (Parts i–viii)
Each part below begins by identifying which identity applies, then shows the substitution and simplification clearly. Study the identity tag in each problem — it tells you exactly which formula was used and why.
Here a = 3k and b = 4l.
Here a = ax² and b = by². The letters in the identity are just placeholders — substitute the full expressions.
Here a = 7d and b = 9e.
Here a = m² and b = n².
Here a = 3t and b = 9s.
Here a = kl and b = mn.
Here x → 6x, a = 5, and b = 6.
Rewrite: (2b − a)(2b + c) = [2b + (−a)][2b + c]. Here x → 2b, a = −a, b = c.
Question 2 — Evaluate Large Numbers Using Identities
This is the most practical application of algebraic identities: computing squares and products of large numbers without a calculator, by splitting them into a round number ± a small number. This method is widely used in competitive exams and is a favourite question type in CBSE, Telangana, and Andhra Pradesh board papers.
Parts (i) & (ii) — Using Identity 1: (a + b)² ≡ a² + 2ab + b²
= 300² + 2(300)(4) + 4²
= 90,000 + 2,400 + 16
= 500² + 2(500)(9) + 9²
= 250,000 + 9,000 + 81
Parts (iii) & (iv) — Using Identity 2: (a − b)² ≡ a² − 2ab + b²
Numbers just below a round number are best expressed as (round number − small number) to apply the (a − b)² identity.
= 1000² − 2(1000)(8) + 8²
= 1,000,000 − 16,000 + 64
= 800² − 2(800)(1) + 1²
= 640,000 − 1,600 + 1
Parts (v) & (vi) — Using Identity 3: (a + b)(a − b) ≡ a² − b²
When two numbers are equidistant from a round number (one above, one below by the same amount), Identity 3 is the perfect shortcut. The product reduces to just two squared terms.
= 300² − 4²
= 90,000 − 16
= 80² − 3²
= 6,400 − 9
Parts (vii) & (viii) — Using Identity 4: (x + a)(x + b) ≡ x² + (a + b)x + ab
When two numbers both exceed a common round number by different amounts, Identity 4 is the perfect fit. The round number becomes x, and the extra amounts become a and b.
= 100² + (9 + 8)(100) + (9)(8)
= 10,000 + 1,700 + 72
= 200² + (4 + 6)(200) + (4)(6)
= 40,000 + 2,000 + 24
Quick Reference — All Answers at a Glance
| Question | Expression | Identity Used | Answer |
|---|---|---|---|
| Q1 (i) | (3k + 4l)² | (a+b)² | 9k² + 24kl + 16l² |
| Q1 (ii) | (ax² + by²)² | (a+b)² | a²x⁴ + 2abx²y² + b²y⁴ |
| Q1 (iii) | (7d − 9e)² | (a−b)² | 49d² − 126de + 81e² |
| Q1 (iv) | (m² − n²)(m² + n²) | (a−b)(a+b) | m⁴ − n⁴ |
| Q1 (v) | (3t + 9s)(3t − 9s) | (a+b)(a−b) | 9t² − 81s² |
| Q1 (vi) | (kl − mn)(kl + mn) | (a−b)(a+b) | k²l² − m²n² |
| Q1 (vii) | (6x + 5)(6x + 6) | (x+a)(x+b) | 36x² + 66x + 30 |
| Q1 (viii) | (2b − a)(2b + c) | (x+a)(x+b) | 4b² − 2ab + 2bc − ac |
| Q2 (i) | 304² | (a+b)² | 92,416 |
| Q2 (ii) | 509² | (a+b)² | 259,081 |
| Q2 (iii) | 992² | (a−b)² | 984,064 |
| Q2 (iv) | 799² | (a−b)² | 638,401 |
| Q2 (v) | 304 × 296 | (a+b)(a−b) | 89,984 |
| Q2 (vi) | 83 × 77 | (a+b)(a−b) | 6,391 |
| Q2 (vii) | 109 × 108 | (x+a)(x+b) | 11,772 |
| Q2 (viii) | 204 × 206 | (x+a)(x+b) | 42,024 |
Common Mistakes to Avoid in Exercise 11.4
- Forgetting the middle term in (a + b)²: Many students write (a + b)² = a² + b², completely omitting the 2ab term. This is the single most common algebra error in Class 8. Always write all three terms: a² + 2ab + b².
- Wrong sign in the middle term of (a − b)²: (a − b)² = a² − 2ab + b² — the last term b² is positive, not negative. Both (a + b)² and (a − b)² end in +b². Only the middle term changes sign.
- Applying (a − b)² when (a + b)(a − b) is needed: In Q1(iv), (m² − n²)(m² + n²) has two different brackets — not the same bracket squared. Always check: are the two brackets identical? If yes, use Identity 1 or 2. If they differ only in the sign of one term, use Identity 3.
- Incorrect squaring of compound expressions: In Q1(i), (3k)² = 9k², not 3k². And (4l)² = 16l², not 4l². Square both the coefficient and the variable.
- Sign error in Q1(viii) when using Identity 4: In (2b − a)(2b + c), the constant in the first bracket is −a, not a. Always rewrite subtraction as addition of a negative before matching to the identity: [2b + (−a)][2b + c].
- Choosing the wrong round number in Q2: For 992², choosing 990 instead of 1000 works but is harder. Always pick the nearest multiple of a power of 10 (100, 1000, etc.) to make mental arithmetic easiest.
What Exercise 11.4 Prepares You For
The four identities introduced in Exercise 11.4 are arguably the most important shortcuts in all of secondary school algebra. In Class 9 Polynomials, you will add two more identities — (a + b + c)² and (a + b)³ — but those too are built from the same binomial multiplication principles you have now mastered.
The "evaluate using identities" skill from Question 2 appears directly in CBSE, Telangana, and Andhra Pradesh board exams at Class 8, 9, and 10 levels, and in competitive entrance exams. Once you know that 204 × 196 = (200 + 4)(200 − 4) = 40000 − 16 = 39984, you can answer such questions in under ten seconds. This same technique is the foundation for Factorisation, which runs in reverse: given a² − b², you factorise it back into (a + b)(a − b).