(2a-1)x^2+ax+5=0

解

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

x = \frac{- a + a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}, x = \frac{- a - a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}

解答ステップ

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

解くとthe二次式

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

簡素化

a^{2} - 4 (2 a - 1) \cdot 5 : a^{2} - 20 (2 a - 1)

x_{1, 2} = \frac{- a \pm a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}

解を分離する

x_{1} = \frac{- a + a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}, x_{2} = \frac{- a - a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}

x = \frac{- a + a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}

x = \frac{- a - a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}

二次equationの解:

x = \frac{- a + a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}, x = \frac{- a - a ^{2} - 20 ( 2 a - 1 )}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}