What is the answer to 2x^2+9x-i=0 ?

The answer to 2x^2+9x-i=0 is x=0.00273…+0.11097…i,x=-4.50273…-0.11097…i

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2x^2+9x-i=0

2 x^{2} + 9 x - i = 0

x = 0.00273 \dots + 0.11097 \dots i, x = - 4.50273 \dots - 0.11097 \dots i

Solution steps

2 x^{2} + 9 x - i = 0

Substitute

x = a + bi

2 (a + bi)^{2} + 9 (a + bi) - i = 0

Expand

2 (a + bi)^{2} + 9 (a + bi) - i : (2 a^{2} - 2 b^{2} + 9 a) + i (- 1 + 9 b + 4 ab)

(2 a^{2} - 2 b^{2} + 9 a) + i (- 1 + 9 b + 4 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 a^{2} - 2 b^{2} + 9 a) + i (- 1 + 9 b + 4 ab) = 0 + 0 i

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

[2 a^{2} - 2 b^{2} + 9 a = 0 - 1 + 9 b + 4 ab = 0]

[2 a^{2} - 2 b^{2} + 9 a = 0 - 1 + 9 b + 4 ab = 0] : (a = 0.00273 \dots, a = - 4.50273 \dots, b = 0.11097 \dots b = - 0.11097 \dots)

Substitute back

x = a + bi

x = 0.00273 \dots + 0.11097 \dots i, x = - 4.50273 \dots - 0.11097 \dots i

z^{2} = (1 + i 3)^{2}

\frac{1 + ai}{6 + bi} = 1 + 5 i

0 = \frac{1 + λi}{1 - λi}

z^{2} - (1 - 2 i) z + (2 - 16 i) = 0

(8 + 9 i) y - (3 + 6 i) x = 16 - 3 i

What is the answer to 2x^2+9x-i=0 ?
The answer to 2x^2+9x-i=0 is x=0.00273…+0.11097…i,x=-4.50273…-0.11097…i