+129839 Hint : the key to solving this kind of problem - if it can be solved - is to express the irrational part of the problem as 2*sqrt(a * b) and then finding two numbers, a and b, that sum to the integer part under the root
Also, if the answer can be factored, the larger term in each pair of factors must come first...........
sqrt (49 - 20sqrt(6) ) =
sqrt (49 - 2sqrt (600) ) =
sqrt ( 49 - 2 sqrt (24 *25) ) =
sqrt 24 + 25 - 2sqrt (24 * 25) ) =
sqrt ( 24 - 2sqrt (24 * 25) + 25) =
sqrt (24 - 2sqrt (24)*sqrt(25) + 25)
And the expression inside the parentheses can be factored as :
[ sqrt (24) - sqrt (25) ] [ sqrt (24) - sqrt (25) ] = [ the larger term must come first]
[sqrt (25) - sqrt(24)] [sqrt (25) - sqrt (24)]
[ sqrt (25) - sqrt (24)]^2
So
sqrt (49 - 20sqrt(6) ) =
sqrt ( [ sqrt (25) - sqrt (24)]^2 ) =
sqrt (25) - sqrt (24) =
5 - 2sqrt(6)
