4x^2-4a^2x+(a^4-b^4)=0

解答

4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0

x = \frac{a ^{2} + b ^{2}}{2}, x = \frac{a ^{2} - b ^{2}}{2}

求解步骤

4 x^{2} - 4 a^{2} x + (a^{4} - b^{4}) = 0

整理后得

4 x^{2} - 4 a^{2} x + a^{4} - b^{4} = 0

使用求根公式求解

x_{1, 2} = \frac{- ( - 4 a ^{2} ) \pm ( - 4 a ^{2} ) ^{2} - 4 \cdot 4 ( a ^{4} - b ^{4} )}{2 \cdot 4}

化简

(- 4 a^{2})^{2} - 4 \cdot 4 (a^{4} - b^{4}) : 4 b^{2}

x_{1, 2} = \frac{- ( - 4 a ^{2} ) \pm 4 b ^{2}}{2 \cdot 4}

将解分隔开

x_{1} = \frac{- ( - 4 a ^{2} ) + 4 b ^{2}}{2 \cdot 4}, x_{2} = \frac{- ( - 4 a ^{2} ) - 4 b ^{2}}{2 \cdot 4}

x = \frac{- ( - 4 a ^{2} ) + 4 b ^{2}}{2 \cdot 4} : \frac{a ^{2} + b ^{2}}{2}

x = \frac{- ( - 4 a ^{2} ) - 4 b ^{2}}{2 \cdot 4} : \frac{a ^{2} - b ^{2}}{2}

二次方程组的解是:

x = \frac{a ^{2} + b ^{2}}{2}, x = \frac{a ^{2} - b ^{2}}{2}

so l v e f or x, p^{2} + q^{2} = x^{2} + y^{2}

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

so l v e f or x, f (1 + x) (x - 5) = y

4 x^{2} - 2 (a^{2} + b^{2}) x + a^{2} b^{2} = 0

so l v e f or x, f (x + 5) = x^{2} + 4 x + 3