x2 + 10ix – 21 = 0
Given: x2 + 10ix – 21 = 0
x2 + 10ix – 21 × 1 = 0
We know, i2 = –1 ⇒ 1 = –i2
By substituting 1 = –i2 in the above equation, we get
x2 + 10ix – 21(–i2) = 0
x2 + 10ix + 21i2 = 0
x2 + 3ix + 7ix + 21i2 = 0
x(x + 3i) + 7i(x + 3i) = 0
(x + 3i) (x + 7i) = 0
x + 3i = 0 or x + 7i = 0
x = –3i or –7i
The roots of the given equation are –3i, –7i
Answered by Aaryan | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
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