What are the solutions to the equation a(b-c)x^2+b(c-a)x+c(a-b)=0 ?

The solutions to the equation a(b-c)x^2+b(c-a)x+c(a-b)=0 are

Find the zeros of a(b-c)x^2+b(c-a)x+c(a-b)=0

The zeros of a(b-c)x^2+b(c-a)x+c(a-b)=0 are

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

x = \frac{- b c + ab + a ^{2} b ^{2} + 4 a ^{2} c ^{2} - 4 a ^{2} b c - 4 ab c ^{2} + 2 a b ^{2} c + b ^{2} c ^{2}}{2 ( ab - a c )}, x = \frac{- b c + ab - a ^{2} b ^{2} + 4 a ^{2} c ^{2} - 4 a ^{2} b c - 4 ab c ^{2} + 2 a b ^{2} c + b ^{2} c ^{2}}{2 ( ab - a c )}; ab - a c \neq = 0

Solution steps

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

Expand

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(b c - ab)^{2} - 4 (ab - a c) (a c - b c) : a^{2} b^{2} + 4 a^{2} c^{2} - 4 a^{2} b c - 4 ab c^{2} + 2 a b^{2} c + b^{2} c^{2}

x_{1, 2} = \frac{- ( b c - ab ) \pm a ^{2} b ^{2} + 4 a ^{2} c ^{2} - 4 a ^{2} b c - 4 ab c ^{2} + 2 a b ^{2} c + b ^{2} c ^{2}}{2 ( ab - a c )}; ab - a c \neq = 0

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- b c + ab + a ^{2} b ^{2} + 4 a ^{2} c ^{2} - 4 a ^{2} b c - 4 ab c ^{2} + 2 a b ^{2} c + b ^{2} c ^{2}}{2 ( ab - a c )}, x = \frac{- b c + ab - a ^{2} b ^{2} + 4 a ^{2} c ^{2} - 4 a ^{2} b c - 4 ab c ^{2} + 2 a b ^{2} c + b ^{2} c ^{2}}{2 ( ab - a c )}; ab - a c \neq = 0

2 x^{2} = 4 x + 1

2 x^{2} + 22 x = 0

(2 x - 1) (x - 5) = 0

- 11 = 0.039 x^{2} - 0.9675 x

x^{2} + 12 x + 35 = 0

What are the solutions to the equation a(b-c)x^2+b(c-a)x+c(a-b)=0 ?
The solutions to the equation a(b-c)x^2+b(c-a)x+c(a-b)=0 are
Find the zeros of a(b-c)x^2+b(c-a)x+c(a-b)=0
The zeros of a(b-c)x^2+b(c-a)x+c(a-b)=0 are