(a + b)² = a² + 2ab + b²
√a² = |a|
√(6 + 2√5) = √(5 + 2√5 + 1) = √(√5² + 2*1*√5 + 1²) = √(√5 + 1)² = |√5 + 1| = { √5 + 1 > 0} = √5 + 1 = 1 + √5
(a + b)² = a² + 2ab + b²
√a² = |a|
√(6 + 2√5) = √(5 + 2√5 + 1) = √(√5² + 2*1*√5 + 1²) = √(√5 + 1)² = |√5 + 1| = { √5 + 1 > 0} = √5 + 1 = 1 + √5