What is the answer to ix^2+8x-7i=0 ?

The answer to ix^2+8x-7i=0 is x=i,x=7i

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ix^2+8x-7i=0

i x^{2} + 8 x - 7 i = 0

x = i, x = 7 i

Solution steps

i x^{2} + 8 x - 7 i = 0

Substitute

x = a + bi

i (a + bi)^{2} + 8 (a + bi) - 7 i = 0

Expand

i (a + bi)^{2} + 8 (a + bi) - 7 i : (- 2 ab + 8 a) + i (- 7 + a^{2} - b^{2} + 8 b)

(- 2 ab + 8 a) + i (- 7 + a^{2} - b^{2} + 8 b) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(- 2 ab + 8 a) + i (- 7 + a^{2} - b^{2} + 8 b) = 0 + 0 i

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

[- 2 ab + 8 a = 0 - 7 + a^{2} - b^{2} + 8 b = 0]

[- 2 ab + 8 a = 0 - 7 + a^{2} - b^{2} + 8 b = 0] : (a = 0, a = 0, b = 1 b = 7)

Substitute back

x = a + bi

x = i, x = 7 i

z - \frac{1}{z} = 3 + 3 i

x^{2} + (i - 10) x + 23 - 11 i = 0

i (x) = 2000 x - 2 x^{2}

r = \frac{v}{i}

z^{2} (3 - i) + 5 (1 - i) z + 2 + 16 i = 0