Solve the following quadratic equations:
ix2 – 4x – 4i = 0
ix2 – 4x – 4i = 0
Given ix2 – 4x – 4i = 0
⇒ ix2 + 4x(–1) – 4i = 0
We have i2 = –1
By substituting –1 = i2 in the above equation, we get
ix2 + 4xi2 – 4i = 0
⇒ i(x2 + 4ix – 4) = 0
⇒ x2 + 4ix – 4 = 0
⇒ x2 + 4ix + 4(–1) = 0
⇒ x2 + 4ix + 4i2 = 0 [∵ i2 = –1]
⇒ x2 + 2ix + 2ix + 4i2 = 0
⇒ x(x + 2i) + 2i(x + 2i) = 0
⇒ (x + 2i)(x + 2i) = 0
⇒ (x + 2i)2 = 0
⇒ x + 2i = 0
∴ x = –2i (double root)
Thus, the roots of the given equation are –2i and –2i.