Question 173445
Let's go at this straight up and see what happens:


Let one of the numbers be {{{x}}} and the other one be {{{y}}}, so we know that:


{{{x-y=11}}} and {{{sqrt(x)-sqrt(y)=1}}}.


Yuck!  You can see right off that those radicals are going to make this messy.


So let's try another tack.


Let's keep {{{x}}} and {{{y}}} as the desired variables, but let's define two more variables, thus:


Let {{{t=sqrt(x)}}} and {{{u=sqrt(y)}}}.  That means that {{{x=t^2}}}  and {{{y=u^2}}}.


Now we can write:


1.  {{{t^2-u^2=11}}} and


2.  {{{t-u=1}}}


Solve Eq. 2 for {{{t}}}, {{{t=u+1}}} and substitute this expression for {{{t}}} in Eq. 1.


{{{(u+1)^2-u^2=11}}}


Expand the binomial and collect terms:


{{{u^2+2u+1-u^2=11}}}


{{{2u+1=11}}}


{{{2u=10}}}


{{{u=5}}}


Since we know that {{{t-u=1}}}, {{{t=6}}}


But {{{x=t^2=36}}} and {{{y=u^2=25}}}


Hence, 36 and 25 are the two numbers.


Check:


{{{36-25=11}}}


{{{sqrt(36)-sqrt(25)=6-5=1}}}


Answer checks.