Logo
Overview

Identity 1A: Square of a Sum

When we multiply a number by itself, we get a square. If that number is a sum (a+b)(a+b), we are finding the area of a square with side length (a+b)(a+b).

(a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

Geometric Proof

Imagine a large square with side a+ba+b. We can split it into 4 sections:

  1. A square of area a2a^2.
  2. A square of area b2b^2.
  3. Two identical rectangles of area a×ba \times b.
a²ababb²abab

Identity 1B: Square of a Difference

(a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2

Derivation: We can derive this from Identity 1A by replacing bb with (−b)(-b): (a+(−b))2=a2+2a(−b)+(−b)2(a + (-b))^2 = a^2 + 2a(-b) + (-b)^2 =a2−2ab+b2= a^2 - 2ab + b^2

Identity 1C: Difference of Squares

This is a very useful pattern for fast calculation. (a+b)(a−b)=a2−b2(a + b)(a - b) = a^2 - b^2

Why does this work? Let’s expand it using the distributive property: (a+b)(a−b)=a(a−b)+b(a−b)(a + b)(a - b) = a(a - b) + b(a - b) =a2−ab+ab−b2= a^2 - ab + ab - b^2 =a2−b2= a^2 - b^2 (The middle terms −ab-ab and +ab+ab cancel out!)

Note

History Note: The Indian mathematician Sridharacharya (750 CE) used a modified form of this identity: a2=(a+b)(a−b)+b2a^2 = (a+b)(a-b) + b^2 to calculate squares quickly.