Question 32069
{{{ sqrt(x)= sqrt( x+2 )-1 }}}
I, too, would move the 1 to the left first.
{{{ sqrt(x)+1= sqrt(x+2) }}}
Square both sides
{{{ (sqrt(x)+1)^2= (sqrt(x+2))^2 }}}
on the right, the square cancels the root
{{{ (sqrt(x)+1)^2 =x+2 }}}
{{{ (sqrt(x)+1)(sqrt(x)+1)=x+2 }}}
Foil the left side
F : {{{ sqrt(x) * sqrt(x) = x }}}
O : {{{ sqrt(x) * 1 = sqrt(x) }}}
I : {{{ 1 * sqrt(x) = sqrt(x) }}}
L : {{{ 1 * 1 = 1 }}}
Bring it all together
{{{ x + 2(sqrt(x))+ 1=x+2 }}}
Move the x and the 1 to the right by subtracting
{{{ 2(sqrt(x))= 1 }}}
divide by 2
{{{ sqrt(x)= 1/2 }}}
square both sides again
{{{ (sqrt(x))^2= (1/2)^2 }}}
{{{ x = 1/4 }}}
Plug the 1/4 into the original equation to check
{{{ sqrt(x)= sqrt( x+2 )-1 }}}
{{{ sqrt(1/4)= sqrt( (1/4)+2 )-1 }}}
{{{ sqrt(1/4)= sqrt(9/4)-1 }}}
{{{ 1/2 = (3/2) -1 }}}
{{{ 1/2 = 1/2 }}}
YES!!!
So x does equal 1/4