"Perfect square" radicals? Reducing radicals-within-radicals?

I guess there are all sorts of general equations you

could use.

One of my favourites is √(a ± b√c) = √x ± √y

which I stumbled across many decades ago

while trying to find the sin, cos and tan of

'unusual' angles.

e.g. sin 75º = (1/2)√(2 + √3) which can be

'simplified' to (√2 + √6) / 4.

In my calculations, I got that first value, while the

second one came from a book, and that piqued

my interest.

Just very simply, with √(a ± b√c) = √x ± √y,

square both sides: a ± b√c = x + y ± 2√(xy).

Equating similar parts gives:

a = x + y

b√c = 2√(xy), or, b^2c = 4xy

From a = x + y, y = a - x

Substituting into the other equation gives:

b^2c = 4x(a - x), from which we get:

4x^2 - 4ax + b^c = 0

Solving gives:

x = [a ± √(a^2 - b^2c)] / 2

Because of symmetry, one of these values

can be x and the other one y.

Getting an answer with fractions shouldn't

hurt too much (as with sin 75º), but the main

point is that a^2 - b^2c must be a square

number.

Notice with your equation -

√(3 - 2√2) = √2 - 1,

if you relate it to √(a - b√c),

then a^2 - b^2c = 3^2 - 2^2(2) = 9 - 8 = 1

which is a square number.

And if x = [a + √(a^2 - b^2c)] / 2,

and y = [a - √(a^2 - b^2c)] / 2,

then x = [3 + √1] / 2 = 2

and y = [3 - √1] / 2 = 1.

So, √(3 - 2√2) = √2 - √1 = √2 - 1.

EDIT:

Here's some recreational fun where you can

make up your own numbers.

Choose any 3 numbers. Call them b, m and n.

e.g. let b = 3, m = 2, n = 4.

Now calculate : a = b(m + n) = 18

Then calculate : c = n(n + 2m) = 32

Then calculate : x = (a + bm) / 2 = 12

Then calculate : y = (a - bm) / 2 = 6

Then we have for :

√(a ± b√c) = √x ± √y,

√(18 ± 3√32) = √12 ± √6

If you don't want fractions involved, then you

have to make sure when choosing numbers

that 'bm' and 'bn' are both even or both odd.

This is something I discovered too.

If you have something of the form:

(a + b) + 2√(ab) = a + 2√(ab) + b

It will resolve into a perfect square:

(√a + √b)²

Now that may be obvious with the letters, but if you have numbers it may not always be obvious. For example:

10 + 2√21 = (7 + 3) + 2√(7*3) = 7 + 2√(7*3) + 3

= (√7 + √3)²

or

14 + 8√3 = 14 + 2*4√3 = 14 + 2√(4²*3) = 14 + 2√(16*3)

= 14 + 2√48 = (8 + 6) + 2√(8*6) = 8 + 2√(8*6) + 6

= (√8 + √6)² = (2√2 + √6)²

Good question. I saw the same problem. I did a search of the wikipedia articles on nested radicals and exact trigonometric constants, and found this:

"In general nested radicals cannot be reduced.

But if for √(a+b√c), R = √(a²-b²c) is rational, and both d = ±√((a±R)/2) and e = ±√((a±R/(2c)) are rational with the appropriate choice of the four ± signs, then √(a+b√c) = d + e√c"

I think I know the reason why this is so. Assume that the nested radical may be written as the sum of a rational number and another radical. Assuming the number under the radical is squarefree, the radical will persist after the squaring, so exactly the same radical must appear under the square root. So indeed, we must have that √(a+b√c) = d+e√c. Squaring both sides, we find that:

a+b√c = d² + e²c + 2de√c

Equating rational and irrational parts:

a = d²+e²c

b = 2de

Solving this system for e:

d = b/(2e)

a = b²/(4e²) + e²c

4ae² = b²+4ce⁴

4ce⁴ - 4ae² + b² = 0

By the quadratic formula:

e² = (4a±√(16a² - 16b²c))/(8c) = (a±√(a²-b²c))/(2c)

e = ±√((a±√(a²-b²c))/(2c))

Similarly, if we instead solve for d:

e = b/(2d)

a = d² + b²c/(4d²)

4ad² = 4d⁴ + b²c

4d⁴ - 4ad² + b²c = 0

d² = (4a±√(16a²-16b²c))/8 = (a±√(a²-b²c))/2

d = ±√((a±√(a²-b²c))/2)

Generally, unless one is doing a lot of work involving nested radicals (such as creating the table of exact trigonometric constants), it's probably easier to derive these formulas from scratch than to try to memorize them.

You can do this: V(3-2V2) = a + bV2

Then you square and you find a and b. There is not always an integer solution for a and b, but many times

3 - 2V2 = a^2 + 2b^2 + 2abV2

So, a^2 + 2b^2 = 3 and -2 = 2ab

ab = -1

a^2 + 2/a^2 = 3

Just by looking you noticed that a^2 = b^2 = 1 is a solution

a^4 - 3a^2 + 2 = 0 ... (1)

a^2 = 1 or 2

a= +/-1 or +/-V2, which is not that useful, but works OK too. If a = V2, then b = -1/V2, so, you obtain your solution too.

If a = 1, then b = -1. This is an introduced root. But a = -1 and b = 1 works OK. a = -V2 and b= 1/V2 is an introduced root too.

So, if the equation (1) has integer or rational roots, or even one like a = V2, b = -1/V2, then you can get what you want to get, if I understood your question right.

Ana

Good question - i've wondered the same in the past. What I've done previously is to assign the left hand side equal to:

x + sqrt(y), square both sides and see if you can solve the equations by setting the LHS radical = RHS radical and the LHS non radical = RHS non radical

This seems to work in math contests where the problems are set to be solvable in a short period of time. Dont know what the general solution is.