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受欢迎的代数 >

Ax^2+2x+3=x^2-Bx+C

解答

A x^{2} + 2 x + 3 = x^{2} - B x + C

解答

x = \frac{- 2 - B + 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}, x = \frac{- 2 - B - 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}; A \neq = 1

求解步骤

A x^{2} + 2 x + 3 = x^{2} - B x + C

将

C

para o lado esquerdo

A x^{2} + 2 x + 3 - C = x^{2} - B x

将

B x

para o lado esquerdo

A x^{2} + 2 x + 3 - C + B x = x^{2}

将

x^{2}

para o lado esquerdo

A x^{2} + 2 x + 3 - C + B x - x^{2} = 0

改写成标准形式

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

(A - 1) x^{2} + (2 + B) x + 3 - C = 0

使用求根公式求解

x_{1, 2} = \frac{- ( 2 + B ) \pm ( 2 + B ) ^{2} - 4 ( A - 1 ) ( 3 - C )}{2 ( A - 1 )}

化简

(2 + B)^{2} - 4 (A - 1) (3 - C) : 4 A C - 12 A + B^{2} + 4 B + 16 - 4 C

x_{1, 2} = \frac{- ( 2 + B ) \pm 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}; A \neq = 1

将解分隔开

x_{1} = \frac{- ( 2 + B ) + 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}, x_{2} = \frac{- ( 2 + B ) - 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}

x = \frac{- ( 2 + B ) + 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )} : \frac{- 2 - B + 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}

x = \frac{- ( 2 + B ) - 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )} : \frac{- 2 - B - 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}

二次方程组的解是:

x = \frac{- 2 - B + 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}, x = \frac{- 2 - B - 4 A C - 12 A + B ^{2} + 4 B + 16 - 4 C}{2 ( A - 1 )}; A \neq = 1

作图

流行的例子

x^{2} + 10 x + 5 = 30

x(x-9)=2+(2x+1)(x-4)

x (x - 9) = 2 + (2 x + 1) (x - 4)

x^{2} - 2 x - 7.88 = 0

solvefor x,x^2y+y^2=10

so l v e f or x, x^{2} y + y^{2} = 10

c^{2} = 1 2^{2} + 1 6^{2}