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受欢迎的代数 >

solvefor j,z=(2-j)(2+3j)

解答

求解

j, z = (2 - j) (2 + 3 j)

解答

j = - \frac{- 2 + - 3 z + 16}{3}, j = \frac{- 3 z + 16 + 2}{3}

求解步骤

z = (2 - j) (2 + 3 j)

展开

(2 - j) (2 + 3 j) : 4 + 4 j - 3 j^{2}

z = 4 + 4 j - 3 j^{2}

交换两边

4 + 4 j - 3 j^{2} = z

将

z

para o lado esquerdo

4 + 4 j - 3 j^{2} - z = 0

改写成标准形式

a x^{2} + b x + c = 0

- 3 j^{2} + 4 j + 4 - z = 0

使用求根公式求解

j_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 3 ) ( 4 - z )}{2 ( - 3 )}

化简

4^{2} - 4 (- 3) (4 - z) : 2 - 3 z + 16

j_{1, 2} = \frac{- 4 \pm 2 - 3 z + 16}{2 ( - 3 )}

将解分隔开

j_{1} = \frac{- 4 + 2 - 3 z + 16}{2 ( - 3 )}, j_{2} = \frac{- 4 - 2 - 3 z + 16}{2 ( - 3 )}

j = \frac{- 4 + 2 - 3 z + 16}{2 ( - 3 )} : - \frac{- 2 + - 3 z + 16}{3}

j = \frac{- 4 - 2 - 3 z + 16}{2 ( - 3 )} : \frac{- 3 z + 16 + 2}{3}

二次方程组的解是:

j = - \frac{- 2 + - 3 z + 16}{3}, j = \frac{- 3 z + 16 + 2}{3}

作图

流行的例子

x^{2} + 108 = 0

- 2 w^{2} + 12 w + 19 = 0

solvefor y^2+x=1,y

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

completesquare y^2+5y-1

co m pl e t es q u a re y^{2} + 5 y - 1

因式分解 x^2+4x=-3

f a c t or x^{2} + 4 x = - 3