(2n-1)x^2+(5n+1)x-3=0

解

(2 n - 1) x^{2} + (5 n + 1) x - 3 = 0

x = \frac{- 5 n - 1 + 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}, x = \frac{- 5 n - 1 - 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}; n \neq = \frac{1}{2}

解答ステップ

(2 n - 1) x^{2} + (5 n + 1) x - 3 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 5 n + 1 ) \pm ( 5 n + 1 ) ^{2} - 4 ( 2 n - 1 ) ( - 3 )}{2 ( 2 n - 1 )}

簡素化

(5 n + 1)^{2} - 4 (2 n - 1) (- 3) : 25 n^{2} + 34 n - 11

x_{1, 2} = \frac{- ( 5 n + 1 ) \pm 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}; n \neq = \frac{1}{2}

解を分離する

x_{1} = \frac{- ( 5 n + 1 ) + 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}, x_{2} = \frac{- ( 5 n + 1 ) - 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}

x = \frac{- ( 5 n + 1 ) + 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )} : \frac{- 5 n - 1 + 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}

x = \frac{- ( 5 n + 1 ) - 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )} : \frac{- 5 n - 1 - 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}

二次equationの解:

x = \frac{- 5 n - 1 + 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}, x = \frac{- 5 n - 1 - 25 n ^{2} + 34 n - 11}{2 ( 2 n - 1 )}; n \neq = \frac{1}{2}