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Simplifying Square Root within Square Root without Calculator

Date: 07/05/2004 at 08:37:07
From: Ahmad
Subject: sqrt under sqrt

I need to find sqrt[5 + sqrt(2)] without using a calculator.  Is there
a general formula for solving a problem like this?

Date: 07/05/2004 at 10:25:19
From: Doctor Vogler
Subject: Re: sqrt under sqrt

Hi Ahmad,

Thanks for writing to Dr Math.  There is a nice little technique that
can unravel the square root of a number like yours.  It doesn't work
with higher roots (like cube roots), but it works well for square 
roots.  It works like this:

  To find sqrt(a + b*sqrt(c)),
  first calculate a^2 - b^2*c.
  If it is not a square, then you can't simplify it.
  If a^2 - b^2*c = q^2, then

    sqrt(a + |b|*sqrt(c)) = sqrt((a+q)/2) + sqrt((a-q)/2)
    sqrt(a - |b|*sqrt(c)) = sqrt((a+q)/2) - sqrt((a-q)/2).

For example, let's try sqrt(8 + sqrt(15)).  In this case, a = 8, 
b = 1, and c = 15.  So:

  a^2 - b^2*c = 8^2 - 1^2*15 = 64 - 15 = 49

  Since 49 is a perfect square, q^2 = 49 and q = 7, and we can
  simplify to sqrt((8+7)/2) + sqrt((8-7)/2) or sqrt(15/2) + sqrt(1/2).

We can simplify that answer a bit, and then we'll check it by squaring
it to see if it really gives us the quantity we originally were trying
to take the square root of.  Simplifying, we get:

  sqrt(15/2) + sqrt(1/2) = sqrt(15)/sqrt(2) + sqrt(1)/sqrt(2)
                         = [sqrt(15) + sqrt(1)] / sqrt(2)
                         = [sqrt(15) + 1] / sqrt(2)

Now let's square that simplified answer to check it:

  {[sqrt(15) + 1] / sqrt(2)}^2  (square the binomial)
  [15 + 2sqrt(15) + 1] / 2
  [16 + 2sqrt(15)] / 2
  8 + sqrt(15)

We started by trying to find the square root of (8 + sqrt(15)), and
our answer does in fact return that value when squared, so we are correct.

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 
Associated Topics:
High School Polynomials
High School Square & Cube Roots

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