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√(3+4√-7)

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Let, (a + ib)2 = 3 + 4√7 i

Now using, (a + b)2 = a2 + b2 + 2ab

⇒ a2 + (bi)2 + 2abi = 3 + 4√7 i

Since i2 = -1

⇒ a2 - b2 + 2abi = 3 + 4√7 i

Now, separating real and complex parts, we get

⇒ a2 - b2 = 3 …………..eq.1

⇒ 2ab = 4√7 …….. eq.2

⇒ a = 2√7/b

Now, using the value of a in eq.1, we get

⇒ \((\frac{2\sqrt7}{b})^2\) – b2 = 3

⇒ 12 – b4 = 3b2

⇒ b4 + 3b2 - 28 = 0

Simplify and get the value of b2, we get,

⇒ - b2 = -7 or b2 = 4

as b is real no. so, b2 = 4

b = 2 or b = - 2

Therefore, a = √7 or a = -√7

Hence the square root of the complex no. is √7 + 2i and -√7 - 2i.

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