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 >

9x^2+8(k-5)x+64=0

解

9 x^{2} + 8 (k - 5) x + 64 = 0

解

x = \frac{4 ( k ^{2} - 10 k - 11 + 5 - k )}{9}, x = \frac{4 ( 5 - k - k ^{2} - 10 k - 11 )}{9}

解答ステップ

9 x^{2} + 8 (k - 5) x + 64 = 0

拡張

9 x^{2} + 8 (k - 5) x + 64 : 9 x^{2} + 8 k x - 40 x + 64

9 x^{2} + 8 k x - 40 x + 64 = 0

標準的な形式で書く

a x^{2} + b x + c = 0

9 x^{2} + (8 k - 40) x + 64 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( 8 k - 40 ) \pm ( 8 k - 40 ) ^{2} - 4 \cdot 9 \cdot 64}{2 \cdot 9}

簡素化

(8 k - 40)^{2} - 4 \cdot 9 \cdot 64 : 8 k^{2} - 10 k - 11

x_{1, 2} = \frac{- ( 8 k - 40 ) \pm 8 k ^{2} - 10 k - 11}{2 \cdot 9}

解を分離する

x_{1} = \frac{- ( 8 k - 40 ) + 8 k ^{2} - 10 k - 11}{2 \cdot 9}, x_{2} = \frac{- ( 8 k - 40 ) - 8 k ^{2} - 10 k - 11}{2 \cdot 9}

x = \frac{- ( 8 k - 40 ) + 8 k ^{2} - 10 k - 11}{2 \cdot 9} : \frac{4 ( k ^{2} - 10 k - 11 + 5 - k )}{9}

x = \frac{- ( 8 k - 40 ) - 8 k ^{2} - 10 k - 11}{2 \cdot 9} : \frac{4 ( 5 - k - k ^{2} - 10 k - 11 )}{9}

二次equationの解:

x = \frac{4 ( k ^{2} - 10 k - 11 + 5 - k )}{9}, x = \frac{4 ( 5 - k - k ^{2} - 10 k - 11 )}{9}

グラフ

人気の例

solvefor t,s=v_{0}*t+a*t^2

so l v e f or t, s = v_{0} \cdot t + a \cdot t^{2}

solvefor y,9y^2=x*(3-x)^2

so l v e f or y, 9 y^{2} = x \cdot (3 - x)^{2}

y (y + 5) = 14

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

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

solvefor x,4x^2+6xy+4y^2=1

so l v e f or x, 4 x^{2} + 6 x y + 4 y^{2} = 1