What are the solutions to the equation bx^2-b=x-b^2x ?

The solutions to the equation bx^2-b=x-b^2x are x= 1/b ,x=-b;\quad b\ne 0

Find the zeros of bx^2-b=x-b^2x

The zeros of bx^2-b=x-b^2x are x= 1/b ,x=-b;\quad b\ne 0

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

x = \frac{1}{b}, x = - b; b \neq = 0

Solution steps

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

Move

b^{2} x

to the left side

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

Move

x

to the left side

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

Write in the standard form

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

b x^{2} + (b^{2} - 1) x - b = 0

Solve with the quadratic formula

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

Simplify

(b^{2} - 1)^{2} - 4 b (- b) : b^{2} + 1

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

Separate the solutions

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

x = \frac{- ( b ^{2} - 1 ) + b ^{2} + 1}{2 b} : \frac{1}{b}

x = \frac{- ( b ^{2} - 1 ) - ( b ^{2} + 1 )}{2 b} : - b

The solutions to the quadratic equation are:

x = \frac{1}{b}, x = - b; b \neq = 0

3 (x^{2} - 7) - 4 = - 7

x^{2} - 31 = 2 x

2 x^{2} + 6 x - 7 = 2

x, 3 x^{2} + 30 x - 5 = 0

5 r^{2} + 24 r - 3 = 2

What are the solutions to the equation bx^2-b=x-b^2x ?
The solutions to the equation bx^2-b=x-b^2x are x= 1/b ,x=-b;\quad b\ne 0
Find the zeros of bx^2-b=x-b^2x
The zeros of bx^2-b=x-b^2x are x= 1/b ,x=-b;\quad b\ne 0