Solve the following quadratic equations:
ix2 – x + 12i = 0
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.