2x^2-5x+2=0,x-2<0

解

2 x^{2} - 5 x + 2 = 0, x - 2 < 0

x = 2, x = \frac{1}{2}

十進法表記

x = 2, x = 0.5

解答ステップ

2 x^{2} - 5 x + 2 = 0

解くとthe二次式

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

(- 5)^{2} - 4 \cdot 2 \cdot 2 = 3

x_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 2}

解を分離する

x_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 2}

x = \frac{- ( - 5 ) + 3}{2 \cdot 2} : 2

x = \frac{- ( - 5 ) - 3}{2 \cdot 2} : \frac{1}{2}

二次equationの解:

x = 2, x = \frac{1}{2}