solvefor x,Ax^2+Cy^2+Dx+Ey+F=0

解答

求解

x, A x^{2} + C y^{2} + D x + E y + F = 0

x = \frac{- D + D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}, x = \frac{- D - D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}; A \neq = 0

求解步骤

A x^{2} + C y^{2} + D x + E y + F = 0

改写成标准形式

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

A x^{2} + D x + C y^{2} + E y + F = 0

使用求根公式求解

x_{1, 2} = \frac{- D \pm D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}

x = \frac{- D + D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}; A \neq = 0

x = \frac{- D - D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}; A \neq = 0

二次方程组的解是:

x = \frac{- D + D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}, x = \frac{- D - D ^{2} - 4 A ( C y ^{2} + E y + F )}{2 A}; A \neq = 0

\frac{( ( 2 x - 1 ) ^{2} )}{3} - \frac{( x ^{2} - 5 )}{8} = \frac{7}{2}

so l v e f or x, 4 x^{2} - (y - 3)^{2} + 4 z^{2} = 4

2.4 + t + 0.1 t^{2} = 0

5 (x^{2} - 1) = 4 (x^{2} + 1)

2 (x - 3) (x + 10) = 0