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x2+3i=0 so x2=-3i x=square root of (-3i)=square root (-3)square root (i) =i(square root(3)([1/(square root (2)](1+i) and i(square root(3)([-1/(square root (2)](1+i) You can multiply through by i if you want, but I left it since it shows you where the answer came from. Note: The square root of i is 1/square root 2(1+i) and -1/square root of 2 (1+i) to see this, try and square them!

Q: How can you solve the equation x2 plus 3i equals 0?

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-4-3i

3x2 = -9 (divide both sides by 3) x2 = -3 (x would have to be the square root of -3) x = ±√-3 x = ±√3i Since you want to solve by factoring: x2 = -3 add 3 to both sides x2 + 3 = 0 x2 - 3i2 = 0 x2 - (√3i)2 = 0 Factor: (x - √3i)(x + √3i) = 0 x - √3i = 0 or x + √3i = 0 x = √3i or x = -√3i

The four roots are:1 + 2i, 1 - 2i, 3i and -3i.

- 2 - 3i

The answer to the question, as stated, is that the other root could be anything. However, if all the coefficients of the quadratic equation are real numbers, then the other root is 1 minus 3i.

(8+6i)-(2+3i)=6+3i 8+6i-2+3i=6+9i

-2 - 3i

[ 2 - 3i ] is.

1/(1+ 3i)

11

(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i

x2 + 9 has no real factors. Its complex factors are (x + 3i) and (x - 3i) where i is the imaginary square root of -1.