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受欢迎的代数 >

solvefor t,2at^2-6at-a+2tb-3b+2c=t^2+3

解答

求解

t, 2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c = t^{2} + 3

解答

t = \frac{3 a - b + 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}, t = \frac{3 a - b - 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}; a \neq = \frac{1}{2}

求解步骤

2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c = t^{2} + 3

将

3

para o lado esquerdo

2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c - 3 = t^{2}

将

t^{2}

para o lado esquerdo

2 a t^{2} - 6 a t - a + 2 t b - 3 b + 2 c - 3 - t^{2} = 0

改写成标准形式

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

(2 a - 1) t^{2} + (- 6 a + 2 b) t - a - 3 b + 2 c - 3 = 0

使用求根公式求解

t_{1, 2} = \frac{- ( - 6 a + 2 b ) \pm ( - 6 a + 2 b ) ^{2} - 4 ( 2 a - 1 ) ( - a - 3 b + 2 c - 3 )}{2 ( 2 a - 1 )}

化简

(- 6 a + 2 b)^{2} - 4 (2 a - 1) (- a - 3 b + 2 c - 3) : 2 11 a^{2} - 4 a c + 5 a + b^{2} - 3 b + 2 c - 3

t_{1, 2} = \frac{- ( - 6 a + 2 b ) \pm 2 11 a ^{2} - 4 a c + 5 a + b ^{2} - 3 b + 2 c - 3}{2 ( 2 a - 1 )}; a \neq = \frac{1}{2}

将解分隔开

t_{1} = \frac{- ( - 6 a + 2 b ) + 2 11 a ^{2} - 4 a c + 5 a + b ^{2} - 3 b + 2 c - 3}{2 ( 2 a - 1 )}, t_{2} = \frac{- ( - 6 a + 2 b ) - 2 11 a ^{2} - 4 a c + 5 a + b ^{2} - 3 b + 2 c - 3}{2 ( 2 a - 1 )}

t = \frac{- ( - 6 a + 2 b ) + 2 11 a ^{2} - 4 a c + 5 a + b ^{2} - 3 b + 2 c - 3}{2 ( 2 a - 1 )} : \frac{3 a - b + 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}

t = \frac{- ( - 6 a + 2 b ) - 2 11 a ^{2} - 4 a c + 5 a + b ^{2} - 3 b + 2 c - 3}{2 ( 2 a - 1 )} : \frac{3 a - b - 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}

二次方程组的解是:

t = \frac{3 a - b + 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}, t = \frac{3 a - b - 11 a ^{2} + 5 a - 4 a c + b ^{2} + 2 c - 3 b - 3}{2 a - 1}; a \neq = \frac{1}{2}

作图

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