Given x² + 1 = 0

We have i² = –1 ⇒ 1 = –i²

By substituting 1 = –i² in the above equation, we get

x² – i² = 0

⇒ (x + i)(x – i) = 0 [∵ a² – b² = (a + b)(a – b)]

⇒ x + i = 0 or x – i = 0

⇒ x = –i or x = i

∴ x = ±i

Thus, the roots of the given equation are ±i.