What is the general rule to simplify a radical expression in a square root?

For example, #sqrt(2sqrt(3)-4)#

1 Answer
George C.
Jul 23, 2018

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 #p, q, r > 0# and #p^2-q^2r# is a perfect square #s^2# then:

#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 #p=-4#, #q=2# and #r=3# we have:

#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#

Related questions
Impact of this question
1416 views around the world
You can reuse this answer
Creative Commons License