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受欢迎的代数 >

solvefor x,x^2+2xi-5=3i

解答

求解

x, x^{2} + 2 x i - 5 = 3 i

解答

x = \frac{3}{2} + \frac{1 - 2}{2} i, x = - \frac{3}{2} - \frac{1 + 2}{2} i

求解步骤

x^{2} + 2 x i - 5 = 3 i

替代

x = a + bi

(a + bi)^{2} + 2 (a + bi) i - 5 = 3 i

展开

(a + bi)^{2} + 2 (a + bi) i - 5 : (a^{2} - b^{2} - 2 b - 5) + i (2 a + 2 ab)

(a^{2} - b^{2} - 2 b - 5) + i (2 a + 2 ab) = 3 i

将

3 i

改写成标准复数形式:

0 + 3 i

(a^{2} - b^{2} - 2 b - 5) + i (2 a + 2 ab) = 0 + 3 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[a^{2} - b^{2} - 2 b - 5 = 0 2 a + 2 ab = 3]

[a^{2} - b^{2} - 2 b - 5 = 0 2 a + 2 ab = 3] : (a = \frac{3}{2}, a = - \frac{3}{2}, b = \frac{1 - 2}{2} b = - \frac{1 + 2}{2})

x = a + bi

代回

x = \frac{3}{2} + \frac{1 - 2}{2} i, x = - \frac{3}{2} - \frac{1 + 2}{2} i

流行的例子

i(x)=(2x+1)/(x-1)

i (x) = \frac{2 x + 1}{x - 1}

∣ z - 2 i ∣ = 4

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

x + (y - 2) i = 3

x^2+(2-2i)x-(4+2x)=0

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