ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

(2x^2-(k+3)x-k+5)/4 =0

解

\frac{2 x ^{2} - ( k + 3 ) x - k + 5}{4} = 0

解

x = \frac{k + 3 + k ^{2} + 14 k - 31}{4}, x = \frac{k + 3 - k ^{2} + 14 k - 31}{4}

解答ステップ

\frac{2 x ^{2} - ( k + 3 ) x - k + 5}{4} = 0

以下で両辺を乗じる:

4

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

解くとthe二次式

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

簡素化

(- k - 3)^{2} - 4 \cdot 2 (- k + 5) : k^{2} + 14 k - 31

x_{1, 2} = \frac{- ( - k - 3 ) \pm k ^{2} + 14 k - 31}{2 \cdot 2}

解を分離する

x_{1} = \frac{- ( - k - 3 ) + k ^{2} + 14 k - 31}{2 \cdot 2}, x_{2} = \frac{- ( - k - 3 ) - k ^{2} + 14 k - 31}{2 \cdot 2}

x = \frac{- ( - k - 3 ) + k ^{2} + 14 k - 31}{2 \cdot 2} : \frac{k + 3 + k ^{2} + 14 k - 31}{4}

x = \frac{- ( - k - 3 ) - k ^{2} + 14 k - 31}{2 \cdot 2} : \frac{k + 3 - k ^{2} + 14 k - 31}{4}

二次equationの解:

x = \frac{k + 3 + k ^{2} + 14 k - 31}{4}, x = \frac{k + 3 - k ^{2} + 14 k - 31}{4}

グラフ

人気の例

(x + 3 a) (x - 4 b) = 0

solvefor x,y^{11}+y^1+y=x^2

so l v e f or x, y^{11} + y^{1} + y = x^{2}

solvefor x,(p+1)x^2-2(p+3)x+3p=0

so l v e f or x, (p + 1) x^{2} - 2 (p + 3) x + 3 p = 0

(z-i)^3=(z+1)^3

(z - i)^{3} = (z + 1)^{3}

2= 1/1-(v^2)/(c^2)

2 = \frac{1}{1} - \frac{v ^{2}}{c ^{2}}