How do you evaluate #\frac { 1} { - \sqrt { 48} }#?
1 Answer
Explanation:
Note that:
#48 = 4^2 * 3#
#sqrt(ab) = sqrt(a)sqrt(b)" "# if#a, b >= 0#
So we find:
#1/(-sqrt(48)) = -1/(sqrt(4^2 * 3))#
#color(white)(1/(-sqrt(48))) = -1/(sqrt(4^2) * sqrt(3))#
#color(white)(1/(-sqrt(48))) = -1/(4sqrt(3))#
#color(white)(1/(-sqrt(48))) = -sqrt(3)/(4sqrt(3)sqrt(3))#
#color(white)(1/(-sqrt(48))) = -sqrt(3)/12#
That is the simplest form.
This is an irrational number. To calculate rational approximations we can use:
#sqrt(a^2+b) = a+b/(2a+b/(2a+b/(2a+b/(2a+...))))#
with
as follows:
#sqrt(48) = sqrt(7^2-1) = 7-1/(14-1/(14-1/(14-1/(14-1/(14-...)))))#
So:
#1/(-sqrt(48)) = -1/(7-1/(14-1/(14-1/(14-1/(14-1/(14-...)))))#
#color(white)(1/(-sqrt(48))) ~~ -1/(7-1/(14-1/14)) = -1/(7-14/195) = -195/1351 ~~ -0.1443375#