What is the answer to solvefor x,(2-i)x+8x^2=0 ?

The answer to solvefor x,(2-i)x+8x^2=0 is x=0,x=-1/4+1/8 i

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solvefor x,(2-i)x+8x^2=0

solve for

x, (2 - i) x + 8 x^{2} = 0

x = 0, x = - \frac{1}{4} + \frac{1}{8} i

Solution steps

(2 - i) x + 8 x^{2} = 0

Substitute

x = a + bi

(2 - i) (a + bi) + 8 (a + bi)^{2} = 0

Expand

(2 - i) (a + bi) + 8 (a + bi)^{2} : (2 a + b + 8 a^{2} - 8 b^{2}) + i (- a + 2 b + 16 ab)

(2 a + b + 8 a^{2} - 8 b^{2}) + i (- a + 2 b + 16 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 a + b + 8 a^{2} - 8 b^{2}) + i (- a + 2 b + 16 ab) = 0 + 0 i

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

[2 a + b + 8 a^{2} - 8 b^{2} = 0 - a + 2 b + 16 ab = 0]

[2 a + b + 8 a^{2} - 8 b^{2} = 0 - a + 2 b + 16 ab = 0] : (a = 0, a = - \frac{1}{4}, b = 0 b = \frac{1}{8})

Substitute back

x = a + bi

x = 0, x = - \frac{1}{4} + \frac{1}{8} i

z = 2 e^{i \frac{2 π}{3}}

z = \frac{- 6 z + 2 i}{i z - 3}

x^{2} + 2 x + 1 - i = 0

y^{2} = 1 - i

∣ z - (1 - i) ∣ = 6

What is the answer to solvefor x,(2-i)x+8x^2=0 ?
The answer to solvefor x,(2-i)x+8x^2=0 is x=0,x=-1/4+1/8 i