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Algebraic Identities intro

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Presentation on theme: "Algebraic Identities intro"— Presentation transcript:

1 Algebraic Identities intro

2 Algebraic Identities Let us learn three important Algebraic Identities which are very useful in solving many problems. We obtain these Identities by multiplying a binomial by another binomial.

3 Identity 1 Let us consider (a + b)2 (a + b)2 = (a + b) (a + b)
= a(a+b) + b(a+b) = a2+ab + ab + b2 = a2+2ab+b2 Thus, (a + b)2 = a2+2ab+b2 x x x x x x

4 Identity 2 Let us consider (a - b)2 (a - b)2 = (a - b) (a - b)
= a(a - b) - b(a - b) = a2 -ab -ab + b2 = a2-2ab+b2 Thus, (a - b)2 = a2- 2ab+b2 x x x x x x

5 Let us consider (a + b) (a - b) (a + b) (a - b) = (a + b) (a - b)
Identity 3 Let us consider (a + b) (a - b) (a + b) (a - b) = (a + b) (a - b) = a(a - b) + b(a - b) = a2 -ab +ab - b2 = a2- b2 Thus, (a + b) (a - b) = a2-b2 x x x x x x

6 Recap (a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2
(a + b) (a - b) = a2 - b2

7 Example 1:- Find the value of (x + 3)2 using suitable identity
Solution: The above algebraic expression is same as following identity (a + b)2 = a2 + 2ab + b2 Where a = x and b = 3 a2 + 2ab + b2 =x2 + 2 x x x (After substituting value of a and b) =x2 + 6x (Ans)

8 Example 2:- Find the value of (y - 5)2 using suitable identity
Solution: The above algebraic expression is same as following identity (a - b)2 = a2 - 2ab + b2 Where a = y and b = 5 a2 - 2ab + b2 =y2 - 2 x y x (After substituting value of a and b) =y y (Ans)

9 Example 3:- Find the value of (p – 5) (p + 5) using suitable identity
Solution: The above algebraic expression is same as following identity (a + b) (a - b) = a2- b2 Where a = p and b = 5 a2- b2 =p2 – 52 (After substituting value of a and b) = p (Ans)

10 Try these Using identity solve (p + 4)2 (x - 7)2 (b + 4) (b - 4)


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