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

解答

d (x) = 5 (x - 3)^{2} + 1

解答

x = \frac{30 + d + d ^{2} + 60 d - 20}{10}, x = \frac{30 + d - d ^{2} + 60 d - 20}{10}

求解步骤

d (x) = 5 (x - 3)^{2} + 1

展开

5 (x - 3)^{2} + 1 : 5 x^{2} - 30 x + 46

d x = 5 x^{2} - 30 x + 46

交换两边

5 x^{2} - 30 x + 46 = d x

将

d x

para o lado esquerdo

5 x^{2} - 30 x + 46 - d x = 0

改写成标准形式

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

5 x^{2} - (30 + d) x + 46 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 30 - d ) \pm ( - 30 - d ) ^{2} - 4 \cdot 5 \cdot 46}{2 \cdot 5}

化简

(- 30 - d)^{2} - 4 \cdot 5 \cdot 46 : d^{2} + 60 d - 20

x_{1, 2} = \frac{- ( - 30 - d ) \pm d ^{2} + 60 d - 20}{2 \cdot 5}

将解分隔开

x_{1} = \frac{- ( - 30 - d ) + d ^{2} + 60 d - 20}{2 \cdot 5}, x_{2} = \frac{- ( - 30 - d ) - d ^{2} + 60 d - 20}{2 \cdot 5}

x = \frac{- ( - 30 - d ) + d ^{2} + 60 d - 20}{2 \cdot 5} : \frac{30 + d + d ^{2} + 60 d - 20}{10}

x = \frac{- ( - 30 - d ) - d ^{2} + 60 d - 20}{2 \cdot 5} : \frac{30 + d - d ^{2} + 60 d - 20}{10}

二次方程组的解是:

x = \frac{30 + d + d ^{2} + 60 d - 20}{10}, x = \frac{30 + d - d ^{2} + 60 d - 20}{10}

作图

流行的例子

v^{''}+2v^'+v=e^{-2t}

v^{^{''}} + 2 v^{^{'}} + v = e^{- 2 t}

y^{''}+y^'-2y=4

y^{^{''}} + y^{^{'}} - 2 y = 4

y^{''}+y=1,y(0)=6,y^'(0)=0

y^{^{''}} + y = 1, y (0) = 6, y^{^{'}} (0) = 0

y^{''}+2y^'+4y=e^{-2t},y(0)=1,y^'(0)=0

y^{^{''}} + 2 y^{^{'}} + 4 y = e^{- 2 t}, y (0) = 1, y^{^{'}} (0) = 0

y^{''}-4y=e^{3x}+xsin(2x)

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