Question 829204
If there is only one square root, you want to isolate it on one side of the equation before squaring.
{{{x-4sqrt(x)=32}}}
{{{-4sqrt(x)=32-x}}} or {{{x-32=4sqrt(x)}}}
Also, if {{{sqrt(x)}}} has to be a real number, it must be that {{{x>=0}}} .
Now we can square both sides, knowing that we may be introducing some exraneous solutions.
{{{(x-32)^2=(4sqrt(x))^2}}}
{{{x^2-64x+1024=4^2*(sqrt(x))^2}}}
{{{x^2-64x+1024=16x}}}
{{{x^2-64x+1024-16x=0}}}
{{{x^2-80x+1024=0}}} 
Factoring, since {{{x^2-80x+1024=(x-16)(x-64)}}} , we transform the equation into
{{{(x-16)(x-64)=0}}} , whose solutions are
{{{x=16}}} and {{{x=64}}}
Is any one of those an extraneous solution?
For {{{x=16}}},
{{{x-4sqrt(x)=16-4sqrt(16)=16-4*4=16-16=0}}} so {{{x=16}}} is not a solution of the original equation. It's an extraneous solution, introduced when we squared both sides.
For {{{x=64}}},
{{{x-4sqrt(x)=64-4sqrt(64)=64-4*8=64-32=32}}}  So {{{highlight(x=64)}}} is a solution to the original equation.