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受欢迎的代数 >

2=(1-2(1)+5)t^2+(2(1)-2(1))t+1

解答

2 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

解答

t = \frac{1}{2}, t = - \frac{1}{2}

+1

十进制

t = 0.5, t = - 0.5

求解步骤

2 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

展开

(1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1 : 4 t^{2} + 1

2 = 4 t^{2} + 1

交换两边

4 t^{2} + 1 = 2

将

2

para o lado esquerdo

4 t^{2} - 1 = 0

使用求根公式求解

t_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 4 ( - 1 )}{2 \cdot 4}

0^{2} - 4 \cdot 4 (- 1) = 4

t_{1, 2} = \frac{- 0 \pm 4}{2 \cdot 4}

将解分隔开

t_{1} = \frac{- 0 + 4}{2 \cdot 4}, t_{2} = \frac{- 0 - 4}{2 \cdot 4}

t = \frac{- 0 + 4}{2 \cdot 4} : \frac{1}{2}

t = \frac{- 0 - 4}{2 \cdot 4} : - \frac{1}{2}

二次方程组的解是:

t = \frac{1}{2}, t = - \frac{1}{2}

作图

流行的例子

- 12 x = 12 x^{2} + 3

solvefor x,3x^2+3y^2+3z^2+2xy-2yz+2xz=0

so l v e f or x, 3 x^{2} + 3 y^{2} + 3 z^{2} + 2 x y - 2 yz + 2 x z = 0

s/2+(s^2)/2-3=0

\frac{s}{2} + \frac{s ^{2}}{2} - 3 = 0

5 a^{2} + 6 a - 147 = - 3

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