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

1.	Solve the following quadratic equations: x2 + 4ix – 4 = 0
Answer» 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.

Answer»

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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