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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!

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

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-4-3i

2+3i, 2-3i

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

- 2 - 3i

-2 - 3i

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

1/(1+ 3i)

[ 2 - 3i ] is.

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.

-2 + 3iThe additive inverse: -(-2 + 3i) = 2 - 3iThe conjugate: -2 - 3i

I assume that "I" is a variable2+5i+6+3i7i+6+3i10i+616i is the answer

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

11

(7 + 3i) + (8 + 9i) = (7 + 8) + (3i + 9i) = (7 + 8) + (3 + 9)i = 15 + 12i Which can also be written as: 15 + 12i = 3(5 + 4i).

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

Plus or minus 3i.

Expressions cannot be solved. Only equations or inequalities may be solved. Also, there is no symbol between 3i and 5.

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

To form the additive inverse, negate all parts of the complex number → 8 + 3i → -8 - 3i The sum of a number and its additive inverse is 0: (8 + 3i) + (-8 - 3i) = (8 + -8) + (3 + -3)i = (8 - 8) + (3 - 3)i = 0 + 0i = 0.

Use the rules of division for complex numbers. Just divide 1 / (4 + 3i). This requires multiplying numerator and denominator of this fraction by (4 - 3i), to get a real number in the denominator.

I would certainly bet on a computer. Try to solve (log z)^-3i. This is log z raised to the -3i power. A computer comes in handy.

s3 - 3s2 + 9s - 27 = s2(s - 3) + 9(s - 3) = (s - 3)(s2 + 9) or = (s - 3)(s2 - (9i2) = (s - 3)(s - 3i)(s + 3i) if you want to find for what values of s the expression equals to zero, then this happens when s - 3 = 0 or s - 3i = 0 or s + 3i = 0 s = 3, ±3i

x2 + 9 = (x + 3i)(x - 3i) The answer is based on a knowledge of imaginary and complex numbers where i2 = -1

0 + 3i

The complex conjugate of 2-3i is 2+3i.

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