Question

For what values of a does the equation a+x= 1+xa^2 have no solution?

Answer 1

We have the equation

`a+x= 1+xa^2`

we leave the terms with x on the left side of the equation and the independent terms on the right side.

`x-xa^2=1-a\\(1-a^2)x=1-a\\`

resolving for x we have that

`x=(1-a)/(1-a^2)`

Since we need that the equation doesn't have solution, then it is necessary that the denominator of x be 0 and this occur when
`1-a^2=0\\1=a^2\\a=\pm1`

Then, for
`a=\pm1` the equation hasn't solution