Question 712396
we have to try to get both terms to be in the same form
i.e. if we factorise 50 we get factor pairs of 
1 and 50
2 and 25
5 and 10
so we can write {{{sqrt(50)}}} as {{{sqrt(2*25)}}}
this is useful because it contains a square number, 25, and also {{{sqrt(2)}}} which is also in the term {{{3sqrt(2)}}}
so now we have
{{{sqrt(2*25) + 3sqrt(2)}}}
we can now take the square root of 25 out of the square root sign giving
{{{5sqrt(2)+3sqrt(2)}}}
adding these together gives the solution that you already knew
{{{8sqrt(2)}}}