Solve x^2+2x+16=0 by completing the square

Question

Solve x^2+2x+16=0 by completing the square

Answer 1

x² + 2x + 16 = 0

x² + 2x = -16 (Subtract 16 on both sides)

x² + 2x + 1 = -16 + 1 (Add (2/2)² on both sides)

(x + 1)² = -15 (Factor using Perfect Square Trinomial and add -16 and 1)

x + 1 = ±√-15 (Square root on both sides)

x = -1 ± i√15 (Subtract 1 on both sides and use i = √-1)

Answer 2

x² + 2x + 1 + 16 - 1 = 0

(x + 1)² + 15 = 0

(x + 1)² = -15 This is not possible over the reals

x + 1 = ±i√15

x = -1 ± i√15