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受欢迎的代数 >

2z+az^2+a=0

解答

2 z + a z^{2} + a = 0

解答

z = \frac{- 1 + - a ^{2} + 1}{a}, z = - \frac{- a ^{2} + 1 + 1}{a}; a \neq = 0

求解步骤

2 z + a z^{2} + a = 0

改写成标准形式

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

a z^{2} + 2 z + a = 0

使用求根公式求解

z_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 aa}{2 a}

化简

2^{2} - 4 aa : 2 1 - a^{2}

z_{1, 2} = \frac{- 2 \pm 2 1 - a ^{2}}{2 a}; a \neq = 0

将解分隔开

z_{1} = \frac{- 2 + 2 1 - a ^{2}}{2 a}, z_{2} = \frac{- 2 - 2 1 - a ^{2}}{2 a}

z = \frac{- 2 + 2 1 - a ^{2}}{2 a} : \frac{- 1 + - a ^{2} + 1}{a}

z = \frac{- 2 - 2 1 - a ^{2}}{2 a} : - \frac{- a ^{2} + 1 + 1}{a}

二次方程组的解是:

z = \frac{- 1 + - a ^{2} + 1}{a}, z = - \frac{- a ^{2} + 1 + 1}{a}; a \neq = 0

作图

流行的例子

x = - x^{2} + 2

5 a^{2} - 10 a = 0

- 3 x^{2} + 6 x - 9 = 0

solvefor x,(x^2)/2+(z^2)/3 =y

so l v e f or x, \frac{x ^{2}}{2} + \frac{z ^{2}}{3} = y

solvefor t,x^2p^2-2txp+2x^2-t^2=0

so l v e f or t, x^{2} p^{2} - 2 t x p + 2 x^{2} - t^{2} = 0