Question 933340
Let the two consecutive positive even integers be {{{n}}} and {{{n+2}}} .
our first equation is {{{sqrt(n(n+2))=2sqrt(30)}}} .
Squaring both sides of the equal sign, we get the equation
{{{n(n+2)=(2sqrt(30))^2}}} ,
which has all the solutions of the original equation.
(It may have other extra solutions, that we call "extraneous solutions",
but we will check out all solutions at the end, and we will eliminate any extraneous solution then).
 
Solving:
{{{n(n+2)=(2sqrt(30))^2}}}
{{{n^2+2n=4*30}}}
{{{n^2+2n=120}}}<--->{{{n^2+2n-120=0}}}
The last equations above are two different versions of the same quadratic equation.
For this particular quadratic equation, there are 3 easy methods to solve it:
"completing the square",
factoring, and
using the "quadratic formula".
It seems to me that in this particular case "completing the square" is the easiest way to the solution, and this is how it goes.
{{{n^2+2n=120}}}
{{{n^2+2n+1=120+1}}}
{{{(n+1)^2=121}}}
{{{(n+1)^2=11^2}}} ---> {{{system(n+1=11,"or",n+1=-11)}}}
Since we are looking for positive integers {{{n}}} and {{{n+2}}} ,
{{{n+1}}} must be a positive integer, so the only solution is
{{{n+1=11}}} ---> {{{n=11-1}}} ---> {{{highlight(n=10)}}} .
So the two consecutive positive even integers area {{{highlight(10)}}} and {{{highlight(12)}}} , and
CHECKING we find that their geometric mean is {{{sqrt(10*12)=sqrt(120)=sqrt(4*30)=sqrt(4)*sqrt(30)=2sqrt(30)}}}
 
OTHER WAYS TO SOLVE THE SAME QUADRATIC EQUATION:

Using the quadratic formula:
The quadratic formula says that the solutions to a quadratic equation
{{{ax^2+bx+c=0}}} , with {{{a<>0}}} are given by
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In the case of {{{n^2+2n-120=0}}} the coefficients are
{{{a=1}}} , {{{b=2}}} , and {{{c=-120}}} , and the solutions will be given by
{{{x = (-2 +- sqrt( 2^2-4*1*(-120) ))/(2*1) }}}
{{{x = (-2 +- sqrt( 4+480))/2}}}
{{{x=(-2 +- 22)/2}}} ---> {{{system(x=(-2+22)/2=20/2=highlight(10),"or",x=(-2- 22)/2=-24/2=-12)}}}
Since {{{-12}}} is not a positive even integer, we eliminate it, and then we check that {{{x=10}}} is a solution to the problem, as done above.

Factoring:
{{{n^2+2n-120=0}}}--->{{{(n+12)(n-10)=0}}}--->{{{system(n+12=0,"or",n-10=0)}}}--->{{{system(n=-12,"or",highlight(n=10))}}}
Since {{{-12}}} is not a positive even integer, we eliminate it, and then we check that {{{x=10}}} is a solution to the problem, as done above.