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受欢迎的代数 >

solvefor q,z(1-q^2)=2(aqx+qy)

解答

求解

q, z (1 - q^{2}) = 2 (a q x + q y)

解答

q = - \frac{2 y + 2 a x + ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}, q = - \frac{2 y + 2 a x - ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}; z \neq = 0

求解步骤

z (1 - q^{2}) = 2 (a q x + q y)

展开

z (1 - q^{2}) : z - z q^{2}

展开

2 (a q x + q y) : 2 a x q + 2 q y

z - z q^{2} = 2 a x q + 2 q y

将

2 q y

para o lado esquerdo

z - z q^{2} - 2 q y = 2 a x q

将

2 a x q

para o lado esquerdo

z - z q^{2} - 2 q y - 2 a x q = 0

改写成标准形式

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

- z q^{2} - (2 y + 2 a x) q + z = 0

使用求根公式求解

q_{1, 2} = \frac{- ( - 2 y - 2 a x ) \pm ( - 2 y - 2 a x ) ^{2} - 4 ( - z ) z}{2 ( - z )}

化简

(- 2 y - 2 a x)^{2} - 4 (- z) z : (- 2 y - 2 a x)^{2} + 4 z^{2}

q_{1, 2} = \frac{- ( - 2 y - 2 a x ) \pm ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 ( - z )}; z \neq = 0

将解分隔开

q_{1} = \frac{- ( - 2 y - 2 a x ) + ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 ( - z )}, q_{2} = \frac{- ( - 2 y - 2 a x ) - ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 ( - z )}

q = \frac{- ( - 2 y - 2 a x ) + ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 ( - z )} : - \frac{2 y + 2 a x + ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}

q = \frac{- ( - 2 y - 2 a x ) - ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 ( - z )} : - \frac{2 y + 2 a x - ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}

二次方程组的解是:

q = - \frac{2 y + 2 a x + ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}, q = - \frac{2 y + 2 a x - ( - 2 y - 2 a x ) ^{2} + 4 z ^{2}}{2 z}; z \neq = 0

作图

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