Answer
$x=0$
Work Step by Step
$\sqrt{1+x}+\sqrt{1-x}=2$
Square both sides of the equation:
$(\sqrt{1+x}+\sqrt{1-x})^{2}=2^{2}$
$(1+x)+2(\sqrt{1+x})(\sqrt{1-x})+(1-x)=4$
$1+x+2\sqrt{1-x^{2}}+1-x=4$
$2\sqrt{1-x^{2}}+2=4$
Take the $2$ to the right side:
$2\sqrt{1-x^{2}}=4-2$
$2\sqrt{1-x^{2}}=2$
Take the $2$ to divide the right side of the equation:
$\sqrt{1-x^{2}}=\dfrac{2}{2}$
$\sqrt{1-x^{2}}=1$
Once again, square both sides:
$(\sqrt{1-x^{2}})^{2}=1^{2}$
$1-x^{2}=1$
Solve for $x$:
$x^{2}=1-1$
$x^{2}=0$
$x=\sqrt{0}$
$x=0$
Check the solution found:
$\sqrt{1+0}+\sqrt{1-0}=2$
$1+1=2$
$2=2$ True
The final answer is $x=0$
