What is the answer to 3x^2+(1-i)x+3i=0 ?

The answer to 3x^2+(1-i)x+3i=0 is x=0.55981…-0.55981…i,x=-0.89314…+0.89314…i

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3x^2+(1-i)x+3i=0

3 x^{2} + (1 - i) x + 3 i = 0

x = 0.55981 \dots - 0.55981 \dots i, x = - 0.89314 \dots + 0.89314 \dots i

Solution steps

3 x^{2} + (1 - i) x + 3 i = 0

Substitute

x = a + bi

3 (a + bi)^{2} + (1 - i) (a + bi) + 3 i = 0

Expand

3 (a + bi)^{2} + (1 - i) (a + bi) + 3 i : (3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab)

(3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(3 a^{2} - 3 b^{2} + a + b) + i (3 - a + b + 6 ab) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[3 a^{2} - 3 b^{2} + a + b = 0 3 - a + b + 6 ab = 0]

[3 a^{2} - 3 b^{2} + a + b = 0 3 - a + b + 6 ab = 0] : (a = 0.55981 \dots, a = - 0.89314 \dots, b = - 0.55981 \dots b = 0.89314 \dots)

Substitute back

x = a + bi

x = 0.55981 \dots - 0.55981 \dots i, x = - 0.89314 \dots + 0.89314 \dots i

x^{2} + (4 - 2 i) x - (1 + i) = 0

x^{2} - 4 x + 40 = 2 - 6 i

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

3 x + 8 i = 12 + 12

(x + y) + (x - y) \cdot i = 5 + 5 \cdot i

What is the answer to 3x^2+(1-i)x+3i=0 ?
The answer to 3x^2+(1-i)x+3i=0 is x=0.55981…-0.55981…i,x=-0.89314…+0.89314…i