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

The solutions to the equation a^2*(x-b)-b^2*(x-a)=x^2*(a-b) are

Find the zeros of a^2*(x-b)-b^2*(x-a)=x^2*(a-b)

The zeros of a^2*(x-b)-b^2*(x-a)=x^2*(a-b) are

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

x = \frac{- b ^{2} + a ^{2} + a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}, x = \frac{- b ^{2} + a ^{2} - a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}; a - b \neq = 0

Solution steps

a^{2} (x - b) - b^{2} (x - a) = x^{2} (a - b)

Expand

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

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

Switch sides

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

Subtract

- b^{2} x + a b^{2}

from both sides

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

Simplify

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

Subtract

a^{2} x - a^{2} b

from both sides

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

Simplify

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

Solve with the quadratic formula

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

Simplify

(b^{2} - a^{2})^{2} - 4 (a - b) (- a b^{2} + a^{2} b) : a^{4} - 4 a^{3} b + 6 a^{2} b^{2} - 4 a b^{3} + b^{4}

x_{1, 2} = \frac{- ( b ^{2} - a ^{2} ) \pm a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}; a - b \neq = 0

Separate the solutions

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

x = \frac{- ( b ^{2} - a ^{2} ) + a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )} : \frac{- b ^{2} + a ^{2} + a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}

x = \frac{- ( b ^{2} - a ^{2} ) - a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )} : \frac{- b ^{2} + a ^{2} - a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}

The solutions to the quadratic equation are:

x = \frac{- b ^{2} + a ^{2} + a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}, x = \frac{- b ^{2} + a ^{2} - a ^{4} - 4 a ^{3} b + 6 a ^{2} b ^{2} - 4 a b ^{3} + b ^{4}}{2 ( a - b )}; a - b \neq = 0

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

a, a^{2} + b^{2} + c^{2} = a + c - \frac{1}{2}

2 x^{2} - 7 = 7 (x - 1)

600 x^{2} - 61 x + 1.5 = 0

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

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