What are the solutions to the equation ax^2+2a=x+2a^2x ?

The solutions to the equation ax^2+2a=x+2a^2x are x=2a,x= 1/a ;\quad a\ne 0

Find the zeros of ax^2+2a=x+2a^2x

The zeros of ax^2+2a=x+2a^2x are x=2a,x= 1/a ;\quad a\ne 0

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

x = 2 a, x = \frac{1}{a}; a \neq = 0

Solution steps

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

Move

2 a^{2} x

to the left side

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

Move

x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 2 a^{2} - 1)^{2} - 4 a \cdot 2 a : (2 a + 1) (2 a - 1)

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 a, x = \frac{1}{a}; a \neq = 0

x = - 2 x^{2} - (b - 8) x + \frac{b}{3}

x^{2} - 49 = 0

\frac{x ^{2}}{25} = 16

s^{2} = - 14 - 3 s

x, x^{2} y = 4 x^{2} + 7 x y + 2 y^{2}

What are the solutions to the equation ax^2+2a=x+2a^2x ?
The solutions to the equation ax^2+2a=x+2a^2x are x=2a,x= 1/a ;\quad a\ne 0
Find the zeros of ax^2+2a=x+2a^2x
The zeros of ax^2+2a=x+2a^2x are x=2a,x= 1/a ;\quad a\ne 0