Question 29890
{{{ sqrt(2x+5) - sqrt(2x-4) = 3 }}} first you must add {{{ sqrt (2x-4) }}} to both sides.
{{{ sqrt(2x+5) = 3 + sqrt(2x-4) }}} Square both sides to get rid of the root on the left
{{{ (sqrt(2x+5))^2 = (3 + sqrt(2x-4))^2 }}} the root and the square on the left cancel each other out
{{{ 2x+5 = (3 + sqrt(2x-4))^2 }}} FOIL the right side
{{{ (3 + sqrt(2x-4))(3 + sqrt(2x-4)) }}}
F: {{{ 3 * 3 = 9 }}}
O: {{{ 3 * sqrt(2x-4) = 3sqrt(2x-4) }}}
I: {{{ 3 * sqrt(2x-4) = 3sqrt(2x-4) }}}
L: {{{ sqrt(2x-4) * sqrt(2x-4) = 2x - 4 }}}This is the right side of the equation.
{{{ 2x+5 = 9 + 3sqrt(2x-4) + 3sqrt(2x-4) + 2x - 4 }}} Combine terms
{{{ 2x+5 = 5 + 2x + 6sqrt(2x-4) }}} subtrace 2x and 5 from both sides to isolate the root.
{{{ 0 = 6sqrt(2x-4) }}} divide both sides by 6
{{{ 0 = sqrt(2x-4) }}} square both sides to get rid of the radicand
{{{ 0^2 = (sqrt(2x-4))^2 }}}
{{{ 0 = 2x-4 }}} add 4 to both sides
{{{ 4 = 2x }}} divide by 2 on both sides
{{{ 2 = x }}}
CHECK YOUR SOLUTION
{{{ sqrt(2x+5) - sqrt(2x-4) = 3 }}}  plug 2 in for x
{{{ sqrt(2(2)+5) - sqrt(2(2)-4) = 3 }}}
{{{ sqrt(4+5) - sqrt(4-4) = 3 }}}
{{{ sqrt(9) - sqrt(0) = 3 }}} root of 9 (assuming only positive roots) is 3, root of 0 is 0
{{{ 3 - 0 = 3 }}} 
{{{ 3 = 3 }}} YES!!