Question

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

Answer 1

Answer:

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

$x=\frac{1-a}{1-a^2}$

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