1+2 b/3+(b^2)/9 =0

解

1 + 2 \frac{b}{3} + \frac{b ^{2}}{9} = 0

b = - 3

解答ステップ

1 + 2 \cdot \frac{b}{3} + \frac{b ^{2}}{9} = 0

以下で両辺を乗じる:

9

b^{2} + 6 b + 9 = 0

解くとthe二次式

b_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot 9}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot 9 = 0

b_{1, 2} = \frac{- 6 \pm 0}{2 \cdot 1}

b = \frac{- 6}{2 \cdot 1}

\frac{- 6}{2 \cdot 1} = - 3

b = - 3

二次equationの解:

b = - 3