(2k-1)x^2+5x-1=0

解

(2 k - 1) x^{2} + 5 x - 1 = 0

x = \frac{- 5 + 8 k + 21}{2 ( 2 k - 1 )}, x = \frac{- 5 - 8 k + 21}{2 ( 2 k - 1 )}; k \neq = \frac{1}{2}

解答ステップ

(2 k - 1) x^{2} + 5 x - 1 = 0

解くとthe二次式

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

簡素化

5^{2} - 4 (2 k - 1) (- 1) : 8 k + 21

x_{1, 2} = \frac{- 5 \pm 8 k + 21}{2 ( 2 k - 1 )}; k \neq = \frac{1}{2}

解を分離する

x_{1} = \frac{- 5 + 8 k + 21}{2 ( 2 k - 1 )}, x_{2} = \frac{- 5 - 8 k + 21}{2 ( 2 k - 1 )}

x = \frac{- 5 + 8 k + 21}{2 ( 2 k - 1 )}; k \neq = \frac{1}{2}

x = \frac{- 5 - 8 k + 21}{2 ( 2 k - 1 )}; k \neq = \frac{1}{2}

二次equationの解:

x = \frac{- 5 + 8 k + 21}{2 ( 2 k - 1 )}, x = \frac{- 5 - 8 k + 21}{2 ( 2 k - 1 )}; k \neq = \frac{1}{2}