Identity: (a + b)^{2} = a^{2} + b^{2} + 2ab
How is this identity obtained? Let's see how. Taking LHS of the identity: (a + b)^{2} This can also be written as: = (a + b)(a + b) Multiply as we do multiplication of two binomials or use F.O.I.L. method and we get: = a(a + b) + b(a + b) = a^{2} + ab + ab + b^{2} On adding like terms we get: = a^{2} + 2ab + b^{2} On rearranging the terms we get: = a^{2} + b^{2} + 2ab Hence, we obtain the identity i.e. (a + b)^{2} = a^{2} + b^{2} + 2ab Following are a few applications to this identity. Example 1: Solve (3x + 2y)^{2} Solution: This proceeds as: The given polynomial (3x + 2y)^{2} represents the identity (a + b)^{2} Where a = 3x and b = 2y On applying values of a and b on the identity i.e. (a + b)^{2} = a^{2} + b^{2} + 2ab we get: (3x + 2y)^{2} = (3x)^{2} + (2y)^{2} + 2(3x)(2y) Expand the exponential forms and we get: = 9x^{2} + 4y^{2} + 2(3x)(2y) On solving multiplication process we get: = 9x^{2} + 4y^{2} + 12xy Hence, (3x + 2y)^{2} = 9x^{2} + 4y^{2} + 12xy Example 2: Solve (6m + 9n)^{2} Solution: This proceeds as: The given polynomial (6m + 9n)^{2} represents the identity (a + b)^{2} Where a = 6m and b = 9n On applying values of a and b on the identity i.e. (a + b)^{2} = a^{2} + b^{2} + 2ab we get: (6m + 9n)^{2} = (6m)^{2} + (9n)^{2} + 2(6m)(9n) Expand the exponential forms and we get: = 36m^{2} + 81n^{2} + 2(6m)(9n) On solving multiplication process we get: = 36m^{2} + 81n^{2} + 108mn Hence, (6m + 9n)^{2} = 36m^{2} + 81n^{2} + 108mn |