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![Find the values of the using the identity (a – b)^2 [Class 9 Ganita] - Exercise Set 4.2](https://cdn.teachoo.com/c3d15c48-c32a-49a2-b024-64562213ae19/slide57.jpg)
Ex 4.2, 2
Last updated at May 15, 2026 by Teachoo
Class 9
Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)
- Consecutive Square Numbers
- Visualising Algebraic Identities
- Exercise Set 4.1
- Factorisation of Algebraic Expressions Using Identities
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Exercise Set 4.2
- More Identities
- Exercise Set 4.3
- Rewriting a2 - b2 Identity
- Factorisation Using Algebra Tiles
- Factorisation by Splitting the Middle term
- Exercise Set 4.4
- Finding New Algebraic Identities
- Simplifying Rational Expressions
- Exercise Set 4.5
- Word Problems
- End-of-Chapter Exercises
Transcript
Ex 4.2, 2 (i) Find the values of the following using the identity (𝑎−𝑏)^2=𝑎^2−2𝑎𝑏+𝑏^2. (i) (79)^2 79^2 =(𝟖𝟎−𝟏)^𝟐 = 80^2+1^2−2 × 80 × 1 = 6,400+1−160 = 6,401−160 = 𝟔,𝟐𝟒𝟏 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 80 & 𝑏 = 1 Ex 4.2, 2 (ii) Find the values of the following using the identity (𝑎−𝑏)^2=𝑎^2−2𝑎𝑏+𝑏^2. (ii) (193)^2 193^2 =(𝟐𝟎𝟎−𝟕)^𝟐 = 200^2+7^2−2 × 200 × 7 = 40,000+49−2,800 = 40,049 −2,800 = 𝟑𝟕,𝟐𝟒𝟗 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 200 & 𝑏 = 7 Ex 4.2, 2 (iii) Find the values of the following using the identity (𝑎−𝑏)^2=𝑎^2−2𝑎𝑏+𝑏^2. (iii) (299)^2 299^2 =(𝟑𝟎𝟎−𝟏)^𝟐 = 300^2+1^2−2 × 300 × 1 = 90,000+1−600 = 90,001−600 = 𝟖𝟗,𝟒𝟎𝟏 (𝑎−𝑏)^2=𝑎^2+𝑏^2−2𝑎𝑏 Putting 𝑎 = 300 & 𝑏 = 1
