What is the general rule to simplify a radical expression in a square root?
For example, #sqrt(2sqrt(3)-4)#
For example,
1 Answer
In your example, we find:
#sqrt(2sqrt(3)-4) = (sqrt(3)-1)i#
Explanation:
I explored this in https://socratic.org/s/aSS7FqaZ finding:
If
#sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2+sqrt(2p-2s)/2#
If these conditions break down then we might expect something like:
#sqrt(p+qsqrt(r)) = sqrt(2p+2s)/2-sqrt(2p-2s)/2#
or:
#sqrt(p+qsqrt(r)) = -sqrt(2p+2s)/2+sqrt(2p-2s)/2#
In particular, with
#s = sqrt(p^2-q^2r) = sqrt(16-12) = 2#
#sqrt(2p+2s)/2 = sqrt(-8+4)/2 = sqrt(-4)/2 = i#
#sqrt(2p-2s)/2 = sqrt(-8-4)/2 = sqrt(-12)/2 = sqrt(3)i#
In your example the radicand is negative, but we can simplify by splitting the radicand into a perfect square:
#sqrt(2sqrt(3)-4) = sqrt(-(3 - 2sqrt(3)+1))#
#color(white)(sqrt(2sqrt(3)-4)) = sqrt(-(sqrt(3)-1)^2)#
#color(white)(sqrt(2sqrt(3)-4)) = (sqrt(3)-1)i#