Q. 2 I3.5( 2 Votes )

# Solve the followi

ix2 – x + 12i = 0

Given ix2 – x + 12i = 0

ix2 + x(–1) + 12i = 0

We have i2 = –1

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

ix2 + xi2 + 12i = 0

i(x2 + ix + 12) = 0

x2 + ix + 12 = 0

x2 + ix – 12(–1) = 0

x2 + ix – 12i2 = 0 [ i2 = –1]

x2 – 3ix + 4ix – 12i2 = 0

x(x – 3i) + 4i(x – 3i) = 0

(x – 3i)(x + 4i) = 0

x – 3i = 0 or x + 4i = 0

x = 3i or –4i

Thus, the roots of the given equation are 3i and –4i.

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