Given: x2 + 2x + 5 = 0
x2 + 2x + 1 + 4 = 0
x2 + 2(x) (1) + 12 + 4 = 0
(x + 1)2 + 4 = 0 [since, (a + b)2 = a2 + 2ab + b2]
(x + 1)2 + 4 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
(x + 1)2 + 4(–i2) = 0
(x + 1)2 – 4i2 = 0
(x + 1)2 – (2i)2 = 0
[By using the formula, a2 – b2 = (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 + i10 + i20 + i30 is a real number?
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