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Solve the following quadratics x2 – 4x + 7 = 0

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Given x2 – 4x + 7 = 0 

⇒ x2 – 4x + 4 + 3 = 0 

⇒ x2 – 2(x)(2) + 22 + 3 = 0 

⇒ (x – 2)2 + 3 = 0 [∵ (a – b)2 = a2 – 2ab + b2

⇒ (x – 2)2 + 3 × 1 = 0 

We have i2 = –1 

⇒ 1 = –i2 

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

(x – 2)2 + 3(–i2) = 0 

⇒ (x – 2)2 – 3i2 = 0

⇒ (x – 2)2 - \((\sqrt{3}i)^2\) = 0

Since a2 – b2 = (a + b)(a – b), we get

Thus, the roots of the given equation are 2 ± \(\sqrt{3}\) i.​​​​​​

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