Question 53082
{{{sqrt(x+1)+sqrt(x-1)=sqrt(2x+1)}}}
{{{((sqrt(x+1)+sqrt(x-1)))^2=(sqrt(2x+1))^2}}}
{{{(sqrt(x+1)+sqrt(x-1))(sqrt(x+1)+sqrt(x-1))=(sqrt(2x+1))^2}}}
{{{(sqrt(x+1))^2+sqrt((x-1)(x+1))+sqrt((x-1)(x+1))+(sqrt(x-1))^2=(sqrt(2x+1))^2}}}
{{{x+1+2sqrt((x-1)(x+1))+x-1=2x+1}}}
{{{2x+0+2sqrt(x^2-1)=2x+1}}}
{{{-2x+2x+2sqrt(x^2-1)=-2x+2x+1}}}
{{{2sqrt(x^2-1)=1}}}
{{{((2sqrt(x^2-1)))^2=1^2}}}
{{{4(x^2-1)=1}}}
{{{4x^2-4=1}}}
{{{4x^2-4+4=1+4}}}
{{{4x^2=5}}}
{{{4x^2/4=5/4}}}
{{{x^2=5/4}}}
{{{sqrt(x^2)=+-sqrt(5/4)}}}
{{{x=+-sqrt(5)/2}}}
{{{x=-sqrt(5)/2}}} is an extraneous solution because when you substitute it back into the original equation you cannot get a real number solution.
Therefore, the only answer you can use is:
{{{x=sqrt(5)/2}}}