What is the answer to i(x)=-2x^2+5x+1 ?

The answer to i(x)=-2x^2+5x+1 is x=-0.18097…-0.03161…i,x=2.68097…-0.46838…i

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i(x)=-2x^2+5x+1

i (x) = - 2 x^{2} + 5 x + 1

x = - 0.18097 \dots - 0.03161 \dots i, x = 2.68097 \dots - 0.46838 \dots i

Solution steps

i (x) = - 2 x^{2} + 5 x + 1

Substitute

x = a + bi

i (a + bi) = - 2 (a + bi)^{2} + 5 (a + bi) + 1

Expand

- b + ia = (- 2 a^{2} + 2 b^{2} + 5 a + 1) + i (5 b - 4 ab)

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

[- b = - 2 a^{2} + 2 b^{2} + 5 a + 1 a = 5 b - 4 ab]

[- b = - 2 a^{2} + 2 b^{2} + 5 a + 1 a = 5 b - 4 ab] : (a = - 0.18097 \dots, a = 2.68097 \dots, b = - 0.03161 \dots b = - 0.46838 \dots)

Substitute back

x = a + bi

x = - 0.18097 \dots - 0.03161 \dots i, x = 2.68097 \dots - 0.46838 \dots i

CM = C (1 + t i)

(3 + 2 i)^{2} y = (x + 3 i)^{2}

(4 x - 18.6 - 4 y) = (6 y) i + (4 x - 2) j

\frac{2 - 3 i}{2 a + bi} = 3 + 2 i

(x + i y)^{2} = 2 x i

What is the answer to i(x)=-2x^2+5x+1 ?
The answer to i(x)=-2x^2+5x+1 is x=-0.18097…-0.03161…i,x=2.68097…-0.46838…i