Exploring Algebraic Identities · Lesson 2 of 8
Visualising Identities
“When symbols feel abstract, squares and rectangles make them confess.”
• Visualise (a+b)² geometrically. • Derive (a+b)²=a²+2ab+b² step by step. • Understand why the identity holds beyond positive lengths. • Distinguish an equation from an identity. • Use the identity to expand expressions and calculate squares.
Visualising (a+b)²
The identity (a+b)² becomes much easier to understand when we connect each term in the expression with an actual area. Instead of treating the identity as something to memorise, we can see where every part of it comes from by using a simple geometric model.
Begin with a square whose side length is a+b. Since the area of a square is side × side, its total area is (a+b)². Now imagine dividing each side into two parts of lengths a and b. These divisions split the large square into smaller regions.
By finding the area of each region and adding them together, we can understand why the expansion of (a+b)² contains the terms a², 2ab, and b².
| Region | Dimensions | Area |
|---|---|---|
| Square | a×a | a² |
| Rectangle | a×b | ab |
| Rectangle | a×b | ab |
| Square | b×b | b² |
The area of the whole square equals the sum of the four parts.
Why Geometry Alone Is Not Enough
In the picture, a and b are lengths and therefore positive. But an algebraic identity must also work for negative and rational values. To establish that, we use distributivity.
Problem
Verify the identity for a=−2 and b=−3.
- 1.Left side: (−5)²=25.
- 2.Right side: 4+2(−2)(−3)+9.
- 3.=4+12+9=25.
- 4.Both sides are equal.
Problem
Verify the identity for a=−2/3 and b=3/4.
- 1.a+b=1/12, so the left side is 1/144.
- 2.a²=4/9, b²=9/16, and 2ab=−1.
- 3.4/9−1+9/16=1/144.
- 4.So the identity also works for these rational values.
Equation and Identity
A statement that two expressions are equal. It may hold only for particular values.
An equation that is true for all allowed values of its variables.
Problem
Compare x²−1=24 and (x+y)²=x²+2xy+y².
- 1.x²−1=24 is true only for x=5 or x=−5, so it is an equation.
- 2.(x+y)²=x²+2xy+y² is true for all x and y, so it is an identity.
Why (a+b)² Is Not a²+b²
The sign of ab decides the comparison. If ab>0, then (a+b)² is larger. If ab<0, it is smaller. If ab=0, the two expressions are equal.
Expanding Binomials
Problem
Use the identity.
- 1.Take a=5x and b=2y.
- 2.(5x+2y)²=(5x)²+2(5x)(2y)+(2y)².
- 3.=25x²+20xy+4y².
Fast Numerical Squaring
Problem
Use (a+b)².
- 1.43=40+3.
- 2.43²=40²+2(40)(3)+3².
- 3.=1600+240+9=1849.
Practice Problems
- Expand (7x+4y)².
- Expand (2.5p+1.5q)².
- Expand (3s/4+8t)².
- Find 64² using the identity.
- Find 105² using the identity.
- Find 205² using the identity.
- Determine when (a+b)² is greater than, less than, or equal to a²+b².
Key Takeaways
• The area model explains every term in (a+b)². • Distributivity proves the identity for all numbers. • An identity is true for all allowed values. • The term 2ab is essential. • Identities simplify both algebra and arithmetic.
Next, we reverse the identity to factor perfect-square expressions and derive (a−b)².