Q. 2 I3.5( 2 Votes )

Solve the followi

Answer :

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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