r^2+5r+4=0

解

r^{2} + 5 r + 4 = 0

r = - 1, r = - 4

解答ステップ

r^{2} + 5 r + 4 = 0

解くとthe二次式

r_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 4 = 3

r_{1, 2} = \frac{- 5 \pm 3}{2 \cdot 1}

解を分離する

r_{1} = \frac{- 5 + 3}{2 \cdot 1}, r_{2} = \frac{- 5 - 3}{2 \cdot 1}

r = \frac{- 5 + 3}{2 \cdot 1} : - 1

r = \frac{- 5 - 3}{2 \cdot 1} : - 4

二次equationの解:

r = - 1, r = - 4