Question

If $x$ is real, then find the solution set of $\sqrt{x+1}+\sqrt{x-1}=1$.

Answer 1

For equation, $\sqrt{x+1}+\sqrt{x-1}=1$,

$x\ge 1\&x\ge -1$ (non-negative terms inside square roots)
$\Rightarrow x\ge 1$ (intersection of sets)

$\Rightarrow \sqrt{x+1}\ge \sqrt{2}>1$
$\Rightarrow \sqrt{x+1}+\sqrt{x-1}>1$ (square roots of real numbers are non-negative)
This contradicts our initial equation for finding a real solution.

$\therefore$ The equation has no real solution.