Question 149817
Let's assume that {{{sqrt(a)+sqrt(b)=sqrt(a+b)}}} for all values of a and b. So now let's just pick arbitrary values for a and b. So let's make {{{a=2}}} and {{{b=3}}}. 



{{{sqrt(a)+sqrt(b)=sqrt(a+b)}}} Start with the given equation.



{{{sqrt(2)+sqrt(3)=sqrt(2+3)}}} Plug in {{{a=2}}} and {{{b=3}}}



{{{sqrt(2)+sqrt(3)=sqrt(5)}}} Add 2 and 3 to get 5



{{{1.41421+1.73205=2.23607}}} Take the square root of 2 to get 1.41421. Take the square root of 3 to get 1.73205. Take the square root of 5 to get 2.23607.



{{{3.14626=2.23607}}} Add 1.41421 and 1.73205 to get 3.14626



Since {{{3.14626<>2.23607}}}, this shows us that {{{sqrt(2)+sqrt(3)<>sqrt(2+3)}}}. 



So this means that {{{sqrt(a)+sqrt(b)<>sqrt(a+b)}}}. There are some values that will make it true. For instance {{{a=0}}} and {{{b=0}}} are one pair of values, {{{a=1}}} and {{{b=0}}} are another. However, in general, the equation {{{sqrt(a)+sqrt(b)=sqrt(a+b)}}} is not true.