J17

Area of square A = b 2
Area of rectangle B = (a – b)b
Area of rectangle C = (a – b)b
Area of square D = (a – b) 2
Total area = (a – b)b + (a – b)b + (a – b) 2 + b 2
= 2(a – b)b + (a – b) 2 + b 2 ……….(ii)
Equating (i) and (ii) gives:
2(a – b)b + (a – b) 2 + b 2 = a 2
Therefore, 2b(a – b) + (a – b) 2 = a 2 – b 2
This means, (a – b)[2b + (a – b)] = a 2 – b 2 , a – b is common.
That is, (a – b)(2b + a – b) = a 2 – b 2 ,
Therefore, (a + b)(a – b) = a 2 – b 2
Expanding, (a + b)(a – b) = a(a – b) + b(a – b)
= a 2 – ab + ab – b 2
= a 2 – b 2
The expression, (a + b)(a – b), is called the difference of two squares.
The three important identities can thus be summarized as:
(a) (a + b) 2 = a 2 + 2ab + b 2
(b) (a – b) 2 = a 2 – 2ab + b 2
(c) (a + b)(a – b) = a 2 – b 2
Example
Expand and simplify:
(a) (x + 4) 2 (b) (x – 5) 2
(c) (x + 3)(x – 3) (d) (2x + 1) 2
(e) (3x + 2)((3x – 2)
Solutions
(a) (x + 4) 2 = (x + 4)(x + 4)
= x(x + 4) + 4(x + 4)
= x 2 + 4x + 4x + 16
= x 2 + 8x + 16
(b) (x – 5) 2 = (x – 5)(x – 5)
= x(x – 5) – 5(x – 5)
= x 2 – 5x – 5x + 25
= x 2 – 10x + 25
(c) (x + 3)(x – 3) = x(x – 3) + 3(x – 3)
= x 2 – 3x + 3x -9
= x 2 – 9
Note: (x 2 – 9 = x 2 – 3 2 )
(d) (2x + 1) 2 = (2x + 1)(2x + 1)
= 2x(2x + 1) + 1(2x + 1)
= 4x 2 + 2x + 2x + 1
= 4x 2 + 4x + 1