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Date: 03/16/2001 at 00:07:07
From: John Tran

Can you please tell me how I would simplify this:

Squareroot(2+Squareroot(3).

I have been working at this equation for a long time and still can
not figure it out.
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```
Date: 03/16/2001 at 10:34:15
From: Doctor Rob
Subject: Re: Radical Question

Thanks for writing to Ask Dr. Math, John.

Unnesting radicals can be tricky. This one isn't too bad.

The key to this kind of problem is to write the polynomial equation
with integer coefficients of which the quantity, let's call it x, is
a root. In this case, it is

x = sqrt(2+sqrt[3]),
x^2 = 2 + sqrt(3),
x^2 - 2 = sqrt(3),
(x^2 - 2)^2 = 3,
x^4 - 4*x^2 + 4 = 3,
x^4 - 4*x^2 + 1 = 0.

Then you have to try to factor it into two quadratic factors of
the form a*x^2 + b*x + c with a and c rational.  If you can do
this, you can unnest the radicals.  In this case, one such
factorization is

(x^2 - sqrt[6]*x + 1)*(x^2 + sqrt[6]*x + 1) = 0.

(You can check that this works. See below for how to find these.)
Descartes' Rule of Signs tells you that your quantity (which is
positive) is a root of the first factor of the first equation.
Using the quadratic formula then gives

x = (sqrt[6]+-sqrt[6-4*1*1])/2,
= (sqrt[6]+-sqrt[2])/2.

Your quantity is the one with the plus sign.

There is one other such factorization,

(x^2 - sqrt[2]*x - 1)*(x^2 + sqrt[2]*x - 1) = 0.

Using the quadratic formula on the first factor shows that your
quantity is a root of it, too.

To find these factorizations, assume they have the form

(a*x^2 + b*x + c)*(A*x^2 + B*x + C) = x^4 - 4*x^2 + 1,

where a, c, A, and C are rational. Now since we are assuming that a is
rational, we can divide the first factor by a and multiply the second
by a. This won't affect the conditions of c, A, and C being rational.
This means that we can assume that a = 1. Now the factorization
equation above implies that

A = 1,
B + b*A = 0,
C + b*B + c*A = -4,
b*C + c*B = 0,
c*C = 1.

The first equation tells us that A = 1. Then B = -b from the second
equation, and C = 1/c from the fifth equation. Then by substitution in
the third and fourth equations, you see that

1/c - b^2 + c = -4,
b/c - c*b = 0,

or

1/c + c = b^2 - 4,
b*(1/c - c) = 0.

Now looking at this last equation, there are two cases, depending on
whether b = 0 or not. If b = 0, then you get the single equation

1/c + c = -4,
c^2 + 4*c + 1 = 0,

which has no rational roots c (since its discriminant 12 is not a
perfect square). Thus b is nonzero, and so 1/c = c, so c^2 = 1, so
c = +-1.  Since c is rational we have solutions. Substituting that
back into the preceding equation, you get

+-2 = b^2 - 4,
b^2 = 4 +- 2 = 6 or 2,
b = +-sqrt(6) or +-sqrt(2).

These four solutions give you the two helpful factorizations, either
of which will allow you to unnest the radicals.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Exponents
High School Polynomials

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