# What is the square root of -24 + 70i?

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### 1 Answer

Let the square root of -24 + 70i be a + ib.

(a + ib)^2 = -24 + 70i

=> a^2 - b^2 + 2abi = -24 + 70i

equating the real and complex parts

a^2 - b^2 = -24

2ab = 70

=> ab = 35

=> b = 35/a

substitute in a^2 - b^2 = -24

=> a^2 - (35/a)^2 = -24

=> a^4 - 35^2 = -24a^2

=> a^4 + 24a^2 - 35^2 = 0

let x = a^2

=> x^2 + 24x - 35^2 = 0

=> x^2 + 49x - 25x - 1225 = 0

=> x(x + 49) - 25(x + 49) = 0

=> (x - 25)(x + 49) = 0

=> x = 25 and x = -49

but x = a^2

a^2 = 25 => a = 5 and -5

a^2 = -49 gives complex values of a but a is a real number, so we ignore this root.

a = 5 , b = 35/5 = 7

a = -5 , b = 35/-5 = -7

**The required square root is 5 + 7i and -5 - 7i**