Question 11339
you have done THE classic error when squaring. You cannot square each term...you square EACH SIDE, so you would have:


{{{(3)^2=(2sqrt(x)+x)^2}}} which will be horrible to expand.


The secret is to get the term with the square root in it by itself and THEN square both sides, as follows.


{{{3=2sqrt(x)+x}}}
{{{3-x=2sqrt(x)}}}


Now square both sides, to give {{{(3-x)^2=4x}}}
{{{9-6x+x^2=4x}}}
{{{x^2 - 10x + 9 = 0}}}
(x-9)(x-1) = 0

--> x-9=0 OR x-1=0
so x=9 or x=1


jon.