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人気のある代数 >

(ax^2-5x+1)+(bx-16x^2-9)=(-14x^2-10x+c)

解

(a x^{2} - 5 x + 1) + (b x - 16 x^{2} - 9) = (- 14 x^{2} - 10 x + c)

解

x = \frac{- b - 5 + 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}, x = \frac{- b - 5 - 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}; a \neq = 2

解答ステップ

(a x^{2} - 5 x + 1) + (b x - 16 x^{2} - 9) = (- 14 x^{2} - 10 x + c)

拡張

(a x^{2} - 5 x + 1) + (b x - 16 x^{2} - 9) : a x^{2} - 16 x^{2} + b x - 5 x - 8

a x^{2} - 16 x^{2} + b x - 5 x - 8 = - 14 x^{2} - 10 x + c

c

を左側に移動します

a x^{2} - 16 x^{2} + b x - 5 x - 8 - c = - 14 x^{2} - 10 x

10 x

を左側に移動します

a x^{2} - 16 x^{2} + b x + 5 x - c - 8 = - 14 x^{2}

14 x^{2}

を左側に移動します

a x^{2} - 2 x^{2} + b x + 5 x - c - 8 = 0

標準的な形式で書く

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

(a - 2) x^{2} + (b + 5) x - c - 8 = 0

解くとthe二次式

x_{1, 2} = \frac{- ( b + 5 ) \pm ( b + 5 ) ^{2} - 4 ( a - 2 ) ( - c - 8 )}{2 ( a - 2 )}

簡素化

(b + 5)^{2} - 4 (a - 2) (- c - 8) : 4 a c + 32 a + b^{2} + 10 b - 8 c - 39

x_{1, 2} = \frac{- ( b + 5 ) \pm 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}; a \neq = 2

解を分離する

x_{1} = \frac{- ( b + 5 ) + 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}, x_{2} = \frac{- ( b + 5 ) - 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}

x = \frac{- ( b + 5 ) + 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )} : \frac{- b - 5 + 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}

x = \frac{- ( b + 5 ) - 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )} : \frac{- b - 5 - 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}

二次equationの解:

x = \frac{- b - 5 + 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}, x = \frac{- b - 5 - 4 a c + 32 a + b ^{2} + 10 b - 8 c - 39}{2 ( a - 2 )}; a \neq = 2

グラフ

人気の例

5 x^{2} - 10 x + 9 = 3

-2x^2+11x+7=10+4x

- 2 x^{2} + 11 x + 7 = 10 + 4 x

completesquare x^2+10x-11=0

co m pl e t es q u a re x^{2} + 10 x - 11 = 0

-x^2-11/3 x+2=0

- x^{2} - \frac{11}{3} x + 2 = 0

solvefor x,f=9x^2-5xy+7y^2-120

so l v e f or x, f = 9 x^{2} - 5 x y + 7 y^{2} - 120