z^2+(sqrt(3)+j)z+1=0

解答

z^{2} + (3 + j) z + 1 = 0

z = \frac{- 3 - j + j ^{2} + 2 3 j - 1}{2}, z = \frac{- 3 - j - j ^{2} + 2 3 j - 1}{2}

求解步骤

z^{2} + (3 + j) z + 1 = 0

使用求根公式求解

z_{1, 2} = \frac{- ( 3 + j ) \pm ( 3 + j ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

化简

(3 + j)^{2} - 4 \cdot 1 \cdot 1 : j^{2} + 23 j - 1

z_{1, 2} = \frac{- ( 3 + j ) \pm j ^{2} + 2 3 j - 1}{2 \cdot 1}

将解分隔开

z_{1} = \frac{- ( 3 + j ) + j ^{2} + 2 3 j - 1}{2 \cdot 1}, z_{2} = \frac{- ( 3 + j ) - j ^{2} + 2 3 j - 1}{2 \cdot 1}

z = \frac{- ( 3 + j ) + j ^{2} + 2 3 j - 1}{2 \cdot 1} : \frac{- 3 - j + j ^{2} + 2 3 j - 1}{2}

z = \frac{- ( 3 + j ) - j ^{2} + 2 3 j - 1}{2 \cdot 1} : \frac{- 3 - j - j ^{2} + 2 3 j - 1}{2}

二次方程组的解是:

z = \frac{- 3 - j + j ^{2} + 2 3 j - 1}{2}, z = \frac{- 3 - j - j ^{2} + 2 3 j - 1}{2}

2 x^{2} + 6 x = 36

12 x^{2} + 15 x = 0

(a + 1) x^{2} + (2 a + 1) x - a = 30

so l v e f or x, x^{2} + x - 1 = 0

1.77 \cdot 1 0^{- 6} - 1.77 \cdot 1 0^{- 5} x - x^{2} = 0