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in Quadratic Equations by (32.9k points)
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Solve the following quadratic equations: x2 + 4ix – 4 = 0

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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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