How to solve this equation, when the unknown variable just disappears? This problem

$\sqrt{1-x^2} + \sqrt{3+x^2} = 2$

has the solution $x = 1$ and $x = -1$.
However, I always get stuck like this:

  
*
  
*$1-x^2 + 3+x^2 = 4$
  
*$4 = 4$
  

How do I isolate that darn unknown?

$(a+b)^2\neq a^2+b^2$ in general. In your case $\sqrt{1-x^2}+\sqrt{3+x^2}=2$ leads to $1-x^2+2\sqrt{1-x^2}\sqrt{3+x^2}+3+x^2=4$, which gives you $2\sqrt{1-x^2}\sqrt{3+x^2}=0$. Since $\sqrt{3+x^2}\neq0$ for all real $x$, so $\sqrt{1-x^2}=0$. Hence $x=\pm1$.
