What are the solutions to the equation ax^2-x-a^2x+a=0 ?

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

Find the zeros of ax^2-x-a^2x+a=0

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 5 x - 1 = - 1

M = (x + 1)^{2} + (x - 1)^{2} - 2

x, y = 4 - \frac{( x - 2 ) ^{2}}{4}

x, 4 x^{2} + 36 y^{2} + 9 z^{2} - 1 = 0

8 x^{2} + 3 x = 4 x^{2} - x

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