Q. 2 F4.5( 2 Votes )

# Solve the following quadratic equations:x2 + 4ix – 4 = 0

x2 + 4ix – 4 = 0

Given x2 + 4ix – 4 = 0

x2 + 4ix + 4(–1) = 0

We have i2 = –1

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

x2 + 4ix + 4i2 = 0

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.

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