a² − b² = (a + b)(a − b). This is the difference of squares algebraic identity. It is used to factorise expressions of the form a² − b² into two binomial factors. The identity holds for all real numbers a and b.
a² − b² = (a + b)(a − b).
Proof: (a+b)(a−b) = a² − ab + ab − b² = a² − b².
Used to factorise expressions where both terms are perfect squares.
99 × 101 = (100−1)(100+1) = 100² − 1 = 9999.
a² + b² cannot be factorised over real numbers.
Formula: a² − b² = (a + b)(a − b)
Proof (by expansion): (a + b)(a − b) = a(a − b) + b(a − b) = a² − ab + ab − b² = a² − b² ✓
The middle terms −ab and +ab cancel out.
Examples:
x² − 9 = x² − 3² = (x + 3)(x − 3)
25 − y² = 5² − y² = (5 + y)(5 − y)
4x² − 49 = (2x)² − 7² = (2x + 7)(2x − 7)
100 − 1 = 10² − 1² = (10 + 1)(10 − 1) = 11 × 9 = 99 ✓
The identity helps evaluate products quickly.
Method: ab = ((a+b)/2)² − ((a−b)/2)²
Example 1: 99 × 101 = (100 − 1)(100 + 1) = 100² − 1² = 10000 − 1 = 9999
Example 2: 48 × 52 = (50 − 2)(50 + 2) = 50² − 2² = 2500 − 4 = 2496
Example 3: 97 × 103 = (100 − 3)(100 + 3) = 100² − 3² = 10000 − 9 = 9991
This approach is faster than direct multiplication.
Standard algebraic identities:
Note: a² + b² cannot be factorised over real numbers. (It factors as (a + bi)(a − bi) over complex numbers only.)
Condition for applying a² − b²: Both terms must be perfect squares and the operation must be subtraction.
a² − b² = (a + b)(a − b). It is called the difference of squares identity. Proof: (a+b)(a−b) = a² − ab + ab − b² = a² − b².
x² − 16 = x² − 4² = (x + 4)(x − 4).
98 × 102 = (100 − 2)(100 + 2) = 100² − 2² = 10000 − 4 = 9996.
No, a² + b² cannot be factorised over real numbers. Only a² − b² can be factorised as (a+b)(a−b).
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Evaluate 101 × 99 Using Algebraic Identity
101 × 99 = 9999. Using identity (a+b)(a-b) = a²−b²: (100+1)(100−1) = 100²−1² = 10000−1 = 9999. Step-by-step solution.
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