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Simplifying the Square Root of (a + b*sqrt(c))

Date: 07/29/2004 at 12:44:01
From: Diego
Subject: More advanced simplification of square roots

I want to know how to simplify more advanced square roots, such as
sqrt(11 - 2sqrt(18)).  I know that the answer is 3 - sqrt(2), and that
it is related to the fact that 9*2 = 18 and 9 + 2 = 11, but I don't
understand why.

Thank you very much,

Diego


Date: 07/29/2004 at 17:45:51
From: Doctor Vogler
Subject: Re: More advanced simplification of square roots

Hi Diego,

Thanks for writing to Dr Math.  That's a good question, and it 
interested me enough, when I was not much older than you, to figure
out how it works.  So we start with a number of the form

  X = sqrt( a + b * sqrt(c) )

and we suppose that all squares in c have been factored out into b, so
that we would say that c is "squarefree."  (For example, we would
change sqrt(12) to 2*sqrt(3).)  Now we want to write that as a sum of
square roots of integers (or perhaps rational numbers).  Sometimes it
happens that you will get a new square root, as in

  sqrt( 12 + 4 * sqrt(5) ) = sqrt(2) + sqrt(10)

So the first thing I did was to convince myself that if X could be
written as a sum of square roots of rational numbers, then it would
either be in the form

  X = r + s * sqrt(c)

or in the form

  X = r * sqrt(d) + s * sqrt(c*d)
    = sqrt(d) * ( r + s * sqrt(c) )

I did this by noticing that the square of a sum of square roots like

  r + s * sqrt(c) + t * sqrt(d)

will always end up with square root terms not found in

  a + b * sqrt(c)

I won't go through all the boring details of this, but will instead
get right to the heart of the problem.

We know that if X is a sum of square roots of rational numbers, then

  X = sqrt(d) * ( r + s * sqrt(c) )

either for d=1 (where that term disappears) or for some other d.  And
we want this to equal

  X = sqrt( a + b * sqrt(c) )

So we square both of them and get

  X^2 = d*r^2 + d*s^2*c + 2*d*r*s * sqrt(c)
  X^2 = a + b * sqrt(c)

Now, since c has no rational square root, these two things can only be
equal if the coefficients of the sqrt(c)'s are equal, and the other
parts are equal, so

  d*r^2 + d*s^2*c = a
  2*d*r*s = b

From these two equations, we find the following remarkable fact:

  a^2 - b^2*c = (d*r^2 + d*s^2*c)^2 - (2*d*r*s)^2 * c
       = d^2*r^4 + 2*d^2*r^2*s^2*c + d^2*s^4*c^2 - 4*d^2*r^2*s^2c
       = d^2*r^4 - 2*d^2*r^2*s^2*c + d^2*s^4*c^2
       = (d*r^2 - d*s^2*c)^2

What this means is that if X can be written as a sum of square roots
like we want, then a^2 - b^2*c has to be the square of a rational
number.  This doesn't usually happen, but if it doesn't, then that
means you can stop looking for a square root of that form.

In your case, it IS a square.  Now let's suppose that we calculate

  a^2 - b^2*c = m^2

and find that it is the square of a rational number m.  Then that means

  d*r^2 + d*s^2*c = a
  2*d*r*s = b
  d*r^2 - d*s^2*c = m

Actually, it really means that

  |d*r^2 - d*s^2*c| = m  (those are absolute value bars)

but we will consider the two possibilities separately.  In fact, they
work out almost the same, so I'll just do one of them here, the one I
already mentioned.  The other case results in a sign changing, and
that will appear in my summary.

Now add a+m, and subtract a-m:

  a+m = 2*d*r^2
  a-m = 2*d*s^2*c

So if we divide by 2 and take the square roots, then we will get

  sqrt((a+m)/2) = r * sqrt(d)
  sqrt((a-m)/2) = s * sqrt(c*d)

and these are exactly the two terms we wanted!

So now let me summarize our findings:

Suppose we want to write

  sqrt( a + b * sqrt(c) )

as a sum of square roots of rational numbers.  Then we calculate

  a^2 - b^2*c

and check it it is the square of a rational number.  If its square
root is not rational, then you can't write your square root as a sum
of square roots of rational numbers.  On the other hand, if its square
root is m, so that

  m^2 = a^2 - b^2*c

then if b is positive,

  sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) + sqrt((a-m)/2)

and if b is negative,

  sqrt( a + b * sqrt(c) ) = sqrt((a+m)/2) - sqrt((a-m)/2)

Unfortunately, there isn't such a nice formula when you want higher
roots like cube roots inside of cube roots, so I don't have any advice
for those, but it will solve the problem you gave me.

It also has the remarkable feature that it works when you want the
square root of a sum of four terms.  The following works because the
steps above can be stated in more generality.  Suppose I want to find
the square root of

  42 - 16 * sqrt(3) - 2 * sqrt(5)

First I treat the sqrt(3)'s as rational, and look at

  [42 - 16 * sqrt(3)] - 2 * sqrt(5)

To write this as a sum of two square roots, I would need

  [42 - 16 * sqrt(3)]^2 - 2^2 * 5

to be a square.  So that means that I need the square root of

  [42 - 16 * sqrt(3)]^2 - 2^2 * 5
      = 2512 - 1344 * sqrt(3)

That doesn't look very nice, but we know how to take square roots of
things like that.  If that is a sum of two square roots, then we would
need

  2512^2 - 1344^2 * 3 = 891136

to be a square of an integer.  And so we punch it in our calculator
and hit the square root button.  And voila!  It is!

  891136 = 944^2

So we calculate

  sqrt( 2512 - 1344 * sqrt(3) )
     = sqrt((2512+944)/2) - sqrt((2512-944)/2)
     = 24 * sqrt(3) - 28

Amazing!  We're on a roll!  Now where were we?  We were trying to get
the square root of

  a^2 - b^2*c = [42 - 16 * sqrt(3)]^2 - 2^2 * 5

and we did.  So next we calculate

  (a+m)/2 = ([42 - 16 * sqrt(3)] + [24 * sqrt(3) - 28])/2
          = 7 + 4 * sqrt(3)

  (a-m)/2 = ([42 - 16 * sqrt(3)] - [24 * sqrt(3) - 28])/2
          = 35 - 20 * sqrt(3)

Now we know that our number is

  sqrt( 42 - 16 * sqrt(3) - 2 * sqrt(5) )
     = sqrt((a+m)/2) - sqrt((a-m)/2)
     = sqrt( 7 + 4 * sqrt(3) ) - sqrt( 35 - 20 * sqrt(3) )

So now we just need to take the square roots of those two things. 
We're almost there!

  7^2 - 4^2 * 3 = 1 = 1^2

so m = 1, and

  (a+m)/2 = (7+1)/2 = 8/2 = 4
  (a-m)/2 = (7-1)/2 = 6/2 = 3

and

  sqrt( 7 + 4 * sqrt(3) ) = sqrt(4) + sqrt(3)
                          = 2 + sqrt(3)

Finally,

  35^2 - 20^2 * 3 = 25 = 5^2

so m = 5, and

  (a+m)/2 = (35+5)/2 = 40/2 = 20
  (a-m)/2 = (35-5)/2 = 30/2 = 15

and therefore

  sqrt( 35 - 20 * sqrt(3) ) = sqrt(20) - sqrt(15)
                            = 2 * sqrt(5) - sqrt(3) * sqrt(5)

and we end up with our final answer:

  sqrt( 42 - 16 * sqrt(3) - 2 * sqrt(5) )
     = 2 + sqrt(3) - 2 * sqrt(5) + sqrt(3) * sqrt(5)

Isn't that amazing?  Math is just wonderful.  Be warned:  You have to
be careful of your signs in all of this, but you can always just fall
back on this policy:  If it doesn't work, then try changing the sign.

If you have any questions about this or need more help, please write
back, and I will try to offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Polynomials
High School Square & Cube Roots

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