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受欢迎的代数 >

solvefor k,k^2+R/L k+1/(LC)=0

解答

求解

k, k^{2} + \frac{R}{L} k + \frac{1}{L C} = 0

解答

k = \frac{- CR + C ^{2} R ^{2} - 4 L C}{2 L C}, k = \frac{- CR - C ^{2} R ^{2} - 4 L C}{2 L C}; L C \neq = 0

求解步骤

k^{2} + \frac{R}{L} k + \frac{1}{L C} = 0

找到

L, L C

的最小公倍数:

L C

乘以最小公倍数=

L C

k^{2} L C + \frac{R}{L} k L C + \frac{1}{L C} L C = 0 \cdot L C; L C \neq = 0

化简

L C k^{2} + CR k + 1 = 0; L C \neq = 0

使用求根公式求解

k_{1, 2} = \frac{- CR \pm ( CR ) ^{2} - 4 L C \cdot 1}{2 L C}

化简

(CR)^{2} - 4 L C \cdot 1 : C^{2} R^{2} - 4 L C

k_{1, 2} = \frac{- CR \pm C ^{2} R ^{2} - 4 L C}{2 L C}; L C \neq = 0

将解分隔开

k_{1} = \frac{- CR + C ^{2} R ^{2} - 4 L C}{2 L C}, k_{2} = \frac{- CR - C ^{2} R ^{2} - 4 L C}{2 L C}

k = \frac{- CR + C ^{2} R ^{2} - 4 L C}{2 L C}; L C \neq = 0

k = \frac{- CR - C ^{2} R ^{2} - 4 L C}{2 L C}; L C \neq = 0

二次方程组的解是:

k = \frac{- CR + C ^{2} R ^{2} - 4 L C}{2 L C}, k = \frac{- CR - C ^{2} R ^{2} - 4 L C}{2 L C}; L C \neq = 0

作图

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