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![[Class 9] Let a = –2 and b = –3. Then (a + b) = –5 and (a + b)2 = 25. - Visualising Algebraic Identities](https://cdn.teachoo.com/30d8dd1d-c21e-4013-b811-f7a1dd78707a/slide16.jpg)
Example 2 - Chapter 4 Class 9 - Exploring Algebraic Identities (Ganita Manjari I)
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
- 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
Example 2 Prove that 𝑎^2+2𝑎𝑏+𝑏^2=(𝑎+𝑏)^2 for 𝑎=−2, b=3 𝑎=−2/3, b=3/4 For 𝒂=−𝟐, 𝐛=𝟑 Solving LHS 𝑎^2+2𝑎𝑏+𝑏^2 =(−𝟐)^𝟐+𝟐 × (−𝟐) × (𝟑)+𝟑^𝟐 =4−12+9 =(4+9)−12 =13−12 =𝟏 Solving RHS (a+b)^2 =(−𝟐+𝟑)^𝟐 =1^2 =𝟏 Since LHS = RHS We can say that 𝒂^𝟐+𝟐𝒂𝒃+𝒃^𝟐=(𝒂+𝒃)^𝟐 For 𝒂=−𝟐/𝟑, 𝐛=𝟑/𝟒 Solving LHS 𝑎^2+2𝑎𝑏+𝑏^2 =((−𝟐)/𝟑)^𝟐+𝟐 × (−𝟐) × (𝟑)+𝟑^𝟐 =4−12+9 =(4+9)−12 =13−12 =𝟏 Solving RHS (a+b)^2 =(−𝟐+𝟑)^𝟐 =1^2 =𝟏 Since LHS = RHS We can say that 𝒂^𝟐+𝟐𝒂𝒃+𝒃^𝟐=(𝒂+𝒃)^𝟐 Solving RHS (a+b)^2 =(−𝟐+𝟑)^𝟐 =1^2 =𝟏
