Question

Solve ix^2- x+ 12i = 0​

Answer 1

Answer:

Given as ix2 – x + 12i = 0

ix2 + x(–1) + 12i = 0 [as we know, i2 = –1]

Therefore 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 [Since, 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

Therefore, the roots of the given equation are -4i, 3i