b^2+sqrt(2)b-4=0

解

b^{2} + 2 b - 4 = 0

b = 2, b = - 22

十進法表記

b = 1.41421 \dots, b = - 2.82842 \dots

解答ステップ

b^{2} + 2 b - 4 = 0

解くとthe二次式

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

(2)^{2} - 4 \cdot 1 \cdot (- 4) = 32

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

解を分離する

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

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

b = \frac{- 2 - 3 2}{2 \cdot 1} : - 22

二次equationの解:

b = 2, b = - 22