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人気のある代数 >

solvefor k,f=kx^3+x^2+k^2x+3k^2+11

解

解く

k, f = k x^{3} + x^{2} + k^{2} x + 3 k^{2} + 11

解

k = \frac{- x ^{3} + x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}, k = \frac{- x ^{3} - x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}; x \neq = - 3

解答ステップ

f = k x^{3} + x^{2} + k^{2} x + 3 k^{2} + 11

辺を交換する

k x^{3} + x^{2} + k^{2} x + 3 k^{2} + 11 = f

f

を左側に移動します

k x^{3} + x^{2} + k^{2} x + 3 k^{2} + 11 - f = 0

標準的な形式で書く

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

(x + 3) k^{2} + x^{3} k + x^{2} + 11 - f = 0

解くとthe二次式

k_{1, 2} = \frac{- x ^{3} \pm ( x ^{3} ) ^{2} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}

簡素化

(x^{3})^{2} - 4 (x + 3) (x^{2} + 11 - f) : x^{6} - 4 (x + 3) (x^{2} + 11 - f)

k_{1, 2} = \frac{- x ^{3} \pm x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}; x \neq = - 3

解を分離する

k_{1} = \frac{- x ^{3} + x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}, k_{2} = \frac{- x ^{3} - x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}

k = \frac{- x ^{3} + x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}; x \neq = - 3

k = \frac{- x ^{3} - x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}; x \neq = - 3

二次equationの解:

k = \frac{- x ^{3} + x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}, k = \frac{- x ^{3} - x ^{6} - 4 ( x + 3 ) ( x ^{2} + 11 - f )}{2 ( x + 3 )}; x \neq = - 3

グラフ

人気の例

5 x = 3 x^{2} + 1

6 x - 5 x^{2} = 8

m^{2} = m + 2

2 x (x + 5) = 3 (x - 1)

x^{2} - 4 = 2 x - 1