x^2-2x+a=0

解答

x^{2} - 2 x + a = 0

x = 1 + - a + 1, x = 1 - - a + 1

求解步骤

x^{2} - 2 x + a = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot a}{2 \cdot 1}

化简

(- 2)^{2} - 4 \cdot 1 \cdot a : 2 1 - a

x_{1, 2} = \frac{- ( - 2 ) \pm 2 1 - a}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 2 ) + 2 1 - a}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 1 - a}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 1 - a}{2 \cdot 1} : 1 + - a + 1

x = \frac{- ( - 2 ) - 2 1 - a}{2 \cdot 1} : 1 - - a + 1

二次方程组的解是:

x = 1 + - a + 1, x = 1 - - a + 1

\frac{y ^{2}}{3} = \frac{y ^{2}}{6} + 2

(3 x + 5)^{2} + 5 (3 x + 5) + 6 = 0

- 5 c^{2} + c - 18 = - 6 c^{2}

(2 x - 1) (x + 2) = (x + 3) (x + 1)

y^{2} - 2 y - 1 = 0