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受欢迎的代数 >

4z^2+4(m-1)*z+m^2-3m=0

解答

4 z^{2} + 4 (m - 1) \cdot z + m^{2} - 3 m = 0

解答

z = \frac{- m + 1 + m + 1}{2}, z = \frac{- m + 1 - m + 1}{2}

求解步骤

4 z^{2} + 4 (m - 1) z + m^{2} - 3 m = 0

展开

4 z^{2} + 4 (m - 1) z + m^{2} - 3 m : 4 z^{2} + 4 m z - 4 z + m^{2} - 3 m

4 z^{2} + 4 m z - 4 z + m^{2} - 3 m = 0

改写成标准形式

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

4 z^{2} + (4 m - 4) z + m^{2} - 3 m = 0

使用求根公式求解

z_{1, 2} = \frac{- ( 4 m - 4 ) \pm ( 4 m - 4 ) ^{2} - 4 \cdot 4 ( m ^{2} - 3 m )}{2 \cdot 4}

化简

(4 m - 4)^{2} - 4 \cdot 4 (m^{2} - 3 m) : 4 m + 1

z_{1, 2} = \frac{- ( 4 m - 4 ) \pm 4 m + 1}{2 \cdot 4}

将解分隔开

z_{1} = \frac{- ( 4 m - 4 ) + 4 m + 1}{2 \cdot 4}, z_{2} = \frac{- ( 4 m - 4 ) - 4 m + 1}{2 \cdot 4}

z = \frac{- ( 4 m - 4 ) + 4 m + 1}{2 \cdot 4} : \frac{- m + 1 + m + 1}{2}

z = \frac{- ( 4 m - 4 ) - 4 m + 1}{2 \cdot 4} : \frac{- m + 1 - m + 1}{2}

二次方程组的解是:

z = \frac{- m + 1 + m + 1}{2}, z = \frac{- m + 1 - m + 1}{2}

作图

流行的例子

(k+1)x^2-2(k-1)x+1=0

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

3 t^{2} - \frac{1}{2} = 0

4sqrt(3)*x^2-5x-2sqrt(3)=0

43 \cdot x^{2} - 5 x - 23 = 0

1 3^{2} = 5^{2} + x^{2}, x

a^2+b^2-70=12^2+x^2

a^{2} + b^{2} - 70 = 1 2^{2} + x^{2}