What are the solutions to the equation a(x^2+1)=x(a^2+1) ?

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

Find the zeros of a(x^2+1)=x(a^2+1)

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

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

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

Solution steps

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

Expand

a (x^{2} + 1) : a x^{2} + a

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

Move

x (a^{2} + 1)

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

x = \frac{- ( - a ^{2} - 1 ) - ( 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, f (x^{2} + y^{2} + z^{2} z^{2} - 2 x y) = 0

3 = xx + xx + xx

x, 3 x + 4 y (3 x^{2} + 4 y^{2}) = 0

4 x^{2} - 2 (a^{4} - b^{4}) x + a^{2} b^{2} = 0

\frac{( 6 x ^{2} )}{5} = 4.8

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