Question 904562
{{{sqrt(4x+1)-sqrt(2x+4)=1}}}


add {{{sqrt(2x+4)}}} to both sides of the equation to get:


{{{sqrt(4x+1)}}} = {{{sqrt(2x+4)+1}}}


square both sides of the eqaution to get:


{{{4x+1}}} = {{{2x + 4 + 2*sqrt(2x+4) + 1}}}


combine like terms to get:


{{{4x+1}}} = {{{2x + 5 + 2*sqrt(2x+4)}}}


subtract 2x and subtract 5 from both sides of the equation to get:


{{{4x + 1 - 2x - 5}}} = {{{2 * sqrt(2x+4)}}}


simplify to get:


{{{2x - 4}}} = {{{2 * sqrt(2x + 4)}}}


divide both sides of the equation by 2 to get:


{{{x - 2}}} = {{{sqrt(2x+4)}}}


square both sides of the equation to get:


{{{(x-2)^2}}} = {{{(2x+4)}}}


simplify to get:


{{{x^2 - 4x + 4}}} = {{{2x + 4}}}


subtract 2x and subtract 4 from both sides of the equation to get:


{{{x^2 - 6x}}} = 0


complete the squares to get:


{{{x-3)^2 - 9}}} = {{{0}}}


add 9 to both sides of the equation to get:


{{{x-3)^2}}} = {{{9}}}


take the square root of both sides of the equation to get:


{{{x-3}}} = +/- {{{3}}}


add 3 to both sides of the equation to get:


x = {{{3 + 3}}} or x = {{{3 - 3}}}


that makes x = 6 or x = 0


when x = 6, the original equation becomes {{{sqrt(25) - sqrt(16) = 1}}} which becomes {{{5 - 4 = 1}}} which is true because 1 = 1.


when x = 0, the original equation becomes {{{sqrt(1) - sqrt(4) = 1}}} which becomes {{{1 - 2 = 1}}} which is false because -1 does not equal 1.


your solution is  therefore that x = 6.