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 May 21st, 2009, 02:56 PM #1 Member   Joined: Feb 2008 Posts: 89 Thanks: 0 simplifying sqrt[a + sqrt(b)] [color=#000080]Greetings: Is there a means by which to simplify the likes of sqrt[2 + sqrt(3)]. That is, can we rewrite this with a single radical? Thanks, Rich B.[/color]
 May 21st, 2009, 05:50 PM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: simplifying sqrt[a + sqrt(b)] Generally speaking, no. Sorry!
May 22nd, 2009, 02:11 AM   #3
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Re: simplifying sqrt[a + sqrt(b)]

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 Originally Posted by nikkor180 [color=#000080]Greetings: Is there a means by which to simplify the likes of sqrt[2 + sqrt(3)]. That is, can we rewrite this with a single radical? Thanks, Rich B.[/color]
I am not aware of a general methods, but a little guesswork can help quite often.

For instance you can try to look for $a+b\sqrt{3}$ (or any form which seems plausible) fulfilling $(a+b\sqrt{3})^2=2 + \sqrt3$ and compare coefficients at the roots.

The expression $2 + \sqrt3$ can be simplified as follows: Notice (this is the initial guess) that
$(1+\sqrt3)^2=4+2\sqrt3=2(2+\sqrt3)$
This yields
$\sqrt{2+\sqrt3}=\frac{1+\sqrt3}{\sqrt2}.$

Sometimes you can also use some trigonometric identities. We have
$\sqrt{2+\sqrt3}=2\sqrt{\frac{1+\frac{\sqrt3}2}2}= 2\sqrt{\frac{1+\cos\frac\pi6}2}=2\cos\frac\pi{12}$

Now as $\cos\frac{\pi}{12}=\frac{\sqrt2(\sqrt3+1)}{4},$ see here we get
$\sqrt{2+\sqrt3}=\frac{\sqrt2(\sqrt3+1)}{2},$
which is the same result as before. (Although, I do not see any quick derivation of the formula for $\cos\frac\pi{12}$ that I copied from wikipedia.)

 June 25th, 2009, 03:18 AM #4 Newbie   Joined: Jun 2009 Posts: 20 Thanks: 0 Re: simplifying sqrt[a + sqrt(b)] You can also check whether its minimal polynomial is reducable. $\sqrt{a + \sqrt{b}}$ is a root of the polynomial $f= (X^2-a)^2-b = X^4-2aX^2+a^2-b$ If this polynomial $f$ is irreducable then $f$ is its minimum polynomial and this will not help you to to simplify the expression $\sqrt{a + \sqrt{b}}$. However if $f$ can be reduced to a product of 2 polynomials of degree 2, say $f= g \cdot h$, then $\sqrt{a + \sqrt{b}}$ is either a root of $g$ or $h$. It's then easy to check with the abc-formula wich of the 4 roots correspond with $\sqrt{a + sqrt{b}}$. To give an example: $z= \sqrt{3-2\sqrt{2}}$ is a root of the polynomal $f= (X^2-3)^2-8 = X^4-6X^2+1$ Notice that $f$ can be reduced to $f= g \cdot h = (X^2+2X-1)\cdot (X^2-2X-1)$, so $z$ is either a root of g or h. By checking all roots of g and h with the abc-formula: $-1\pm\sqrt{2}$ (roots of g) and $1\pm sqrt{2}$ (roots of h) we find: $z= \sqrt{3-2\sqrt{2}} = \sqrt{2}-1$. However the sad part, your expression can not be simplified this way Then you have to stick to kompiks method.

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