5x^2+4x+(k+1)=0

解答

5 x^{2} + 4 x + (k + 1) = 0

x = \frac{- 2 + - 5 k - 1}{5}, x = - \frac{- 5 k - 1 + 2}{5}

求解步骤

5 x^{2} + 4 x + (k + 1) = 0

整理后得

5 x^{2} + 4 x + k + 1 = 0

使用求根公式求解

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 ( k + 1 )}{2 \cdot 5}

化简

4^{2} - 4 \cdot 5 (k + 1) : 2 - 5 k - 1

x_{1, 2} = \frac{- 4 \pm 2 - 5 k - 1}{2 \cdot 5}

将解分隔开

x_{1} = \frac{- 4 + 2 - 5 k - 1}{2 \cdot 5}, x_{2} = \frac{- 4 - 2 - 5 k - 1}{2 \cdot 5}

x = \frac{- 4 + 2 - 5 k - 1}{2 \cdot 5} : \frac{- 2 + - 5 k - 1}{5}

x = \frac{- 4 - 2 - 5 k - 1}{2 \cdot 5} : - \frac{- 5 k - 1 + 2}{5}

二次方程组的解是:

x = \frac{- 2 + - 5 k - 1}{5}, x = - \frac{- 5 k - 1 + 2}{5}

3 \cdot x^{2} + 10 x - 83 = 0

so l v e f or a, m^{2} + a^{2} = 0

lo g_{10} (10) (a (a + 3)) = 1

x^{2} - (k + 2) x + (k + 5) = 0

so l v e f or a, a^{2} + b^{2} + 2 (3 a + 2 b) + 13 = 0