MIZORAM Class 9 - Algebraic Identities: Quadratic Videos

a + b = 13

∴ (a + b)² = 169

∴ a² + 2ab + b² = 169 … using (a + b)² = a² + 2ab + b²

∴ 5 + 2ab = 169 … since a² + b² = 5

∴ ab = 82

Since (a − b)² = a² − 2ab + b²

(a − b)² =(2x − 3)² = 4x² - 2 × 2x × 3 + 9 = 4x² - 12x + 9

Let (x² − y²) = a and (x² + y²) = b

∴ (x² − y²)² + 2 (x² − y²)(x² + y²) + (x² + y²)² = a² + 2ab + b²

Since (a + b)² = a² + 2ab + b²

∴ (x² − y² + x² + y²)² = (2x²)² = 4x⁴