x^2+(k+2)x+2k+9=0

解

x^{2} + (k + 2) x + 2 k + 9 = 0

x = \frac{- k - 2 + k ^{2} - 4 k - 32}{2}, x = \frac{- k - 2 - k ^{2} - 4 k - 32}{2}

解答ステップ

x^{2} + (k + 2) x + 2 k + 9 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( k + 2 ) \pm ( k + 2 ) ^{2} - 4 \cdot 1 \cdot ( 2 k + 9 )}{2 \cdot 1}

簡素化

(k + 2)^{2} - 4 \cdot 1 \cdot (2 k + 9) : k^{2} - 4 k - 32

x_{1, 2} = \frac{- ( k + 2 ) \pm k ^{2} - 4 k - 32}{2 \cdot 1}

解を分離する

x_{1} = \frac{- ( k + 2 ) + k ^{2} - 4 k - 32}{2 \cdot 1}, x_{2} = \frac{- ( k + 2 ) - k ^{2} - 4 k - 32}{2 \cdot 1}

x = \frac{- ( k + 2 ) + k ^{2} - 4 k - 32}{2 \cdot 1} : \frac{- k - 2 + k ^{2} - 4 k - 32}{2}

x = \frac{- ( k + 2 ) - k ^{2} - 4 k - 32}{2 \cdot 1} : \frac{- k - 2 - k ^{2} - 4 k - 32}{2}

二次equationの解:

x = \frac{- k - 2 + k ^{2} - 4 k - 32}{2}, x = \frac{- k - 2 - k ^{2} - 4 k - 32}{2}