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受欢迎的代数 >

ax^2+(b-1)x+(c+2)=-3x^2+4x-1

解答

a x^{2} + (b - 1) x + (c + 2) = - 3 x^{2} + 4 x - 1

解答

x = \frac{- b + 5 + - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}, x = \frac{- b + 5 - - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}; a \neq = - 3

求解步骤

a x^{2} + (b - 1) x + (c + 2) = - 3 x^{2} + 4 x - 1

展开

a x^{2} + (b - 1) x + (c + 2) : a x^{2} + b x - x + c + 2

a x^{2} + b x - x + c + 2 = - 3 x^{2} + 4 x - 1

将

1

para o lado esquerdo

a x^{2} + b x - x + c + 3 = - 3 x^{2} + 4 x

将

4 x

para o lado esquerdo

a x^{2} + b x - 5 x + c + 3 = - 3 x^{2}

将

3 x^{2}

para o lado esquerdo

a x^{2} + b x - 5 x + c + 3 + 3 x^{2} = 0

改写成标准形式

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

(a + 3) x^{2} + (b - 5) x + c + 3 = 0

使用求根公式求解

x_{1, 2} = \frac{- ( b - 5 ) \pm ( b - 5 ) ^{2} - 4 ( a + 3 ) ( c + 3 )}{2 ( a + 3 )}

化简

(b - 5)^{2} - 4 (a + 3) (c + 3) : - 4 a c - 12 a + b^{2} - 10 b - 12 c - 11

x_{1, 2} = \frac{- ( b - 5 ) \pm - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}; a \neq = - 3

将解分隔开

x_{1} = \frac{- ( b - 5 ) + - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}, x_{2} = \frac{- ( b - 5 ) - - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}

x = \frac{- ( b - 5 ) + - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )} : \frac{- b + 5 + - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}

x = \frac{- ( b - 5 ) - - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )} : \frac{- b + 5 - - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}

二次方程组的解是:

x = \frac{- b + 5 + - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}, x = \frac{- b + 5 - - 4 a c - 12 a + b ^{2} - 10 b - 12 c - 11}{2 ( a + 3 )}; a \neq = - 3

作图

流行的例子

1440=30x-0.1x^2

1440 = 30 x - 0.1 x^{2}

100+50t-4.9t^2=0

100 + 50 t - 4.9 t^{2} = 0

quadraticformula x^2-4x=0

q u a d r a t i c f or m u l a x^{2} - 4 x = 0

x^{2} + 0.5 x + 1 = 0

5=2(x-2)(x+3)+(x-3)(x+3)

5 = 2 (x - 2) (x + 3) + (x - 3) (x + 3)