# Simplify this expression with nested radical signs

My question is-

Simplify:

$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}$$

-
Your thoughts on how to begin??! – rschwieb Jun 8 '12 at 11:47
Three answers, none mine, and still no up-votes except mine. – Michael Hardy Jun 8 '12 at 12:09

\begin{align} & {}\quad \frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}\\[10pt] & =\frac {1}{\sqrt{ 12-2 \sqrt {35}}} \frac {\sqrt{ 12+2 \sqrt {35}}}{\sqrt{ 12+2 \sqrt {35}}}- \frac {2 }{\sqrt{ 10+2 \sqrt {21}}} \frac {\sqrt{ 10-2 \sqrt {21}}}{\sqrt{ 10-2 \sqrt {21}}}- \frac{1}{\sqrt {8+2\sqrt {15}}}\frac{\sqrt {8-2\sqrt {15}}}{\sqrt {8-2\sqrt {15}}}\\[10pt] & =\frac {\sqrt{ 12+2 \sqrt {35}}}{2}-\frac {\sqrt{ 10-2 \sqrt {21}}}{2}-\frac {\sqrt {8-2\sqrt {15}}}{2}\\[10pt] & =\frac {{\sqrt{ 12+2 \sqrt {35}}}- {\sqrt{ 10-2 \sqrt {21}}}- {\sqrt {8-2\sqrt {15}}}}{2}\\[10pt] & = \frac {{{|\sqrt 5 + \sqrt 7|}}- {{|\sqrt 3 - \sqrt 7|}}- {{|\sqrt 3 - \sqrt 5|}}}{2}=\sqrt 3 \end{align}

-
how did you solve the numerator? – mgh Jun 8 '12 at 11:51
This is not correct. – Phira Jun 8 '12 at 11:54
I noticed, I was editing my answer @Phira. – Gigili Jun 8 '12 at 11:57
This answer is wrong: typing the original expression into a calculator gives $1.732\dots$, but $\sqrt{3} - \sqrt{5}$ is negative. – huon Jun 8 '12 at 12:45
Fun, I have never in my entire life been asked a question which required factorizing $a+b\sqrt{c}$ into a square this way. Probably a good sign of my lack of commutative algebra. – rschwieb Jun 8 '12 at 13:51

The general method for going from a square root expression to a polynomial equation is as follows:

Rationalize the denominators and then set your expression equal to $x$.

Now, isolate one of the square roots on one side of the equation and square everything. Repeating this will eventually remove all square roots and will give you a polynomial for $x$. Then, use estimates for $x$ to decide which root of the polynomial is $x$.

In that particular case, the answer is $x=\sqrt 3$, and we can do much better by recognizing the general structure.

The rationalized expression is actually of the form:

$$\frac{\sqrt{b+c+2\sqrt{bc}}}2-\frac{\sqrt{a+c-2\sqrt{ac}}}2-\frac{\sqrt{a+b-2\sqrt{ab}}}2$$

for $a=3$, $b=5$ and $c=7$.

These are complete squares, so we get:

$$\frac 12 (\sqrt b +\sqrt c +\sqrt a-\sqrt c +\sqrt a -\sqrt b) = \sqrt a=\sqrt 3.$$

-
@Gigili But it is, and I am sure that you will eventually edit your answer again and pretend that you came up with it yourself. But really, be my guest after you took into account the order of the numbers 3,5,7. – Phira Jun 8 '12 at 12:29
Seems your answer is correct, I missed a negative sign. – Gigili Jun 8 '12 at 12:58

Let's follow this elementary way to work out things:

$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}= \frac1{\sqrt{({\sqrt{7}-\sqrt{5}})^2}}-\frac2{\sqrt{({\sqrt{7}+\sqrt{3}})^2}}-\frac1{\sqrt{({\sqrt{5}+\sqrt{3}})^2}}=\frac1{({\sqrt{7}-\sqrt{5}})}-\frac2{({\sqrt{7}+\sqrt{3}})}-\frac1{({\sqrt{5}+\sqrt{3}})}=$$ $$\frac{\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{2} =\sqrt3.$$

The proof is complete.

-
@OP Comparing this with other answers, this shows you can do the rationalization or the "completing the square" steps in either order. – rschwieb Jun 8 '12 at 13:54
It's not a proof. Also, Posting an answer at the time you did while there were other answers with the same idea doesn't seem wise. – Gigili Jun 8 '12 at 14:28

Hint: Rationalize the denominators:

For example:

$$\sqrt{12-2\sqrt{35}}\sqrt{12+2\sqrt{35}}=\sqrt{144-4\cdot35}=\sqrt{4}=2$$

-
Yes i got this but how to solve the numerator? – mgh Jun 8 '12 at 11:53
@meg_1997 Without any more information about how you want this to be simplified, I can only speculate that converting from a sum-of-reciprocals-of-radicals to a sum-of-radicals is the best simplification. Of course, we can't always expect every expression to reduce to something nice. (Also, I'd like to throw in that "solve" and "simplify" are completely different tasks...) – rschwieb Jun 8 '12 at 12:01
Looks like I didn't spot that the expressions under the roots had cute factorizations into squares :) The other answer makes use of this. – rschwieb Jun 8 '12 at 13:50

If you did not notice that each radical denests via $\rm\:a+b+2\sqrt{ab}\, =\, (\sqrt{a}+\sqrt{b})^2\:$ then you could easily calculate this by using this radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}$

Recall $\rm\: w = a + b\sqrt{n}\:$ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2$

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\rm\:a\!+\!b+\sqrt{ab}\:$ has norm $\rm\:(a\!-\!b)^2.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = a\!-\!b\$ yields $\rm\: 2b+\sqrt{ab}\:$

and this has $\rm\ \sqrt{trace}\: =\: 2\sqrt{b},\ \ hence,\ \ \ \color{brown}{dividing\ it\ out}\$ of this yields the sqrt: $\rm\ \sqrt{b}+\sqrt{a}.$