Given: x^{2} + 2x + 5 = 0

x^{2} + 2x + 1 + 4 = 0

x^{2} + 2(x) (1) + 1^{2} + 4 = 0

(x + 1)^{2} + 4 = 0 [since, (a + b)^{2} = a^{2} + 2ab + b^{2}]

(x + 1)^{2} + 4 × 1 = 0

We know, i^{2} = –1 ⇒ 1 = –i^{2}

By substituting 1 = –i^{2} in the above equation, we get

(x + 1)^{2} + 4(–i^{2}) = 0

(x + 1)^{2} – 4i^{2} = 0

(x + 1)^{2} – (2i)^{2} = 0

[By using the formula, a^{2} – b^{2} = (a + b) (a – b)]

(x + 1 + 2i)(x + 1 – 2i) = 0

x + 1 + 2i = 0 or x + 1 – 2i = 0

x = –1 – 2i or x = –1 + 2i

The roots of the given equation are -1+2i, -1-2i

Answered by Aaryan | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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