根 ax^2+bx+c

解答

根

a x^{2} + b x + c

x = \frac{- b + b ^{2} - 4 a c}{2 a}, x = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

求解步骤

a x^{2} + b x + c

根与 X 轴相交

(y = 0)

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

使用求根公式求解

x_{1, 2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}

x = \frac{- b + b ^{2} - 4 a c}{2 a}; a \neq = 0

x = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

二次方程组的解是:

x = \frac{- b + b ^{2} - 4 a c}{2 a}, x = \frac{- b - b ^{2} - 4 a c}{2 a}; a \neq = 0

\frac{d y}{d x} + \frac{x}{y} = 0

x y^{^{'}} + 2 y = \frac{3 e ^{- 3 x}}{x}

\frac{\partial}{\partial x} (y^{2} - x^{2} - 10)

\frac{t d y}{d t} = t^{3} + 23 t^{3} y

y^{^{''}} - \frac{2 3}{3} y^{^{'}} + \frac{2}{9} = 0