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受欢迎的代数 >

j(t)=-1/3 t^2+6t-10

解答

j (t) = - \frac{1}{3} t^{2} + 6 t - 10

解答

t = - \frac{- 18 + 3 j + 9 j ^{2} - 108 j + 204}{2}, t = - \frac{- 18 + 3 j - 9 j ^{2} - 108 j + 204}{2}

求解步骤

j (t) = - \frac{1}{3} t^{2} + 6 t - 10

在两边乘以

3

3 j t = - t^{2} + 18 t - 30

交换两边

- t^{2} + 18 t - 30 = 3 j t

将

3 j t

para o lado esquerdo

- t^{2} + 18 t - 30 - 3 j t = 0

改写成标准形式

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

- t^{2} + (18 - 3 j) t - 30 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( 18 - 3 j ) \pm ( 18 - 3 j ) ^{2} - 4 ( - 1 ) ( - 30 )}{2 ( - 1 )}

化简

(18 - 3 j)^{2} - 4 (- 1) (- 30) : 9 j^{2} - 108 j + 204

t_{1, 2} = \frac{- ( 18 - 3 j ) \pm 9 j ^{2} - 108 j + 204}{2 ( - 1 )}

将解分隔开

t_{1} = \frac{- ( 18 - 3 j ) + 9 j ^{2} - 108 j + 204}{2 ( - 1 )}, t_{2} = \frac{- ( 18 - 3 j ) - 9 j ^{2} - 108 j + 204}{2 ( - 1 )}

t = \frac{- ( 18 - 3 j ) + 9 j ^{2} - 108 j + 204}{2 ( - 1 )} : - \frac{- 18 + 3 j + 9 j ^{2} - 108 j + 204}{2}

t = \frac{- ( 18 - 3 j ) - 9 j ^{2} - 108 j + 204}{2 ( - 1 )} : - \frac{- 18 + 3 j - 9 j ^{2} - 108 j + 204}{2}

二次方程组的解是:

t = - \frac{- 18 + 3 j + 9 j ^{2} - 108 j + 204}{2}, t = - \frac{- 18 + 3 j - 9 j ^{2} - 108 j + 204}{2}

作图

流行的例子

solvefor x,x^2-2xy-y^2=6

so l v e f or x, x^{2} - 2 x y - y^{2} = 6

v^{2} - 15 = - 2 v

((y-1)^2)/(49)-((y-1)^2)/(36)=1

\frac{( y - 1 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{36} = 1

x^2-6x+8=-x^2+7x-10

x^{2} - 6 x + 8 = - x^{2} + 7 x - 10

- 3 x^{2} - 7 x + 2 = 0