Not all expressions of this form can be simplified; but if one can, we
would like the square root to have the form
sqrt(a) + sqrt(b) or sqrt(a) - sqrt(b)
where a and b are positive rational numbers, and a > b to give the
positive square root.
If we assume that sqrt(2-sqrt(3)) has this form, then this must be
true:
(sqrt(a) +- sqrt(b))^2 = 2 - sqrt(3)
Expanding, we get
a + b +- 2 sqrt(ab) = 2 - sqrt(3)
Equating the rational and irrational parts of this equation,
a + b = 2
+-2 sqrt(ab) = -sqrt(3)
The second equation here tells us that we will have to choose the
negative sign in order to have a solution, since square roots have to
be positive. If we now square this equation, we have
a + b = 2
4ab = 3
Solving these equations by substituting in the second, we get
4a(2 - a) = 3
4a^2 - 8a - 3 = 0
8 +- sqrt(64 + 48)
a = ------------------ = 1/2 or 3/2
8
and
b = 2 - a = 3/2 or 1/2.
One choice here will give the positive root, the other the negative
root. We take a > b, so that a = 3/2 and b = 1/2. Consequently, our
solution is
sqrt(2 - sqrt(3)) = sqrt(3/2) - sqrt(1/2) = (sqrt(6) - sqrt(2))/2
Looking for information on this technique, which I didn't recall being
taught, I found this page, which together with some rather advanced
techniques includes a quote from an old algebra text about the method:
A slope is found using the formula rise (change in y) over run (change in x). The change in x is 0, and the change in y is -2. The answer is -2/0, but dividing by 0 gives you undefined. Don't believe me? Look at the attached image.
A system of linear equations has one solution when the graphs intersect at a point. A system of linear equations has no solution when the graphs are parallel. A system of linear equations has infinite solutions when the graphs are the exact same line.
