#40sqrt6# is the most reduced form you get. You can't go any further.
Now if you were to approximate #sqrt6#, you can approximate an answer for #40sqrt6#.
#sqrt4 = 2#
#sqrt6 = ?#
#sqrt9 = 3#
If you assume that some #%# between #x_1^2# and #x_2^2# is the same as the #%# between #x_1# and #x_2#, you can say that:
#(6-4)/(9-4)*(3-2)+2 = 2.4# (or #40%# of the way between #2# and #3#)
is the approximate value of #sqrt6#, which would give #40sqrt6 ~~ 40*2.4 = 96#. It's an underestimate though, since what we are really doing is drawing a straight line with a slope of #5# coming from #y = 4# and intersecting at #x = 2# and #x = 3#, and this line is to the left of the actual #x^2# curve at #y = 6#:
graph{(y-x^2)(y + 6 - 5x) = 0 [-1.5, 4, -0.5, 10]}
In actuality it is about #2.45#, and #40sqrt6 ~~ 97.98#.
