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My question is-

Simplify:

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

MJD
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mgh
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5 Answers5

5

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:

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

Phira
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  • @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 08 '12 at 12:29
  • Seems your answer is correct, I missed a negative sign. – Gigili Jun 08 '12 at 12:58
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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.

user 1591719
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  • @OP Comparing this with other answers, this shows you can do the rationalization or the "completing the square" steps in either order. – rschwieb Jun 08 '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 08 '12 at 14:28
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Hint: Rationalize the denominators:

For example:

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

Michael Hardy
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rschwieb
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  • Yes i got this but how to solve the numerator? – mgh Jun 08 '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 08 '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 08 '12 at 13:50
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$$ \begin{align} & {}\quad \frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}\\[10pt] & =\frac {1}{\sqrt{ 12-2 \sqrt {35}}} \frac {\sqrt{ 12+2 \sqrt {35}}}{\sqrt{ 12+2 \sqrt {35}}}- \frac {2 }{\sqrt{ 10+2 \sqrt {21}}} \frac {\sqrt{ 10-2 \sqrt {21}}}{\sqrt{ 10-2 \sqrt {21}}}- \frac{1}{\sqrt {8+2\sqrt {15}}}\frac{\sqrt {8-2\sqrt {15}}}{\sqrt {8-2\sqrt {15}}}\\[10pt] & =\frac {\sqrt{ 12+2 \sqrt {35}}}{2}-\frac {\sqrt{ 10-2 \sqrt {21}}}{2}-\frac {\sqrt {8-2\sqrt {15}}}{2}\\[10pt] & =\frac {{\sqrt{ 12+2 \sqrt {35}}}- {\sqrt{ 10-2 \sqrt {21}}}- {\sqrt {8-2\sqrt {15}}}}{2}\\[10pt] & = \frac {{{|\sqrt 5 + \sqrt 7|}}- {{|\sqrt 3 - \sqrt 7|}}- {{|\sqrt 3 - \sqrt 5|}}}{2}=\sqrt 3 \end{align} $$

Gigili
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  • how did you solve the numerator? – mgh Jun 08 '12 at 11:51
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    This is not correct. – Phira Jun 08 '12 at 11:54
  • I noticed, I was editing my answer @Phira. – Gigili Jun 08 '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 08 '12 at 12:45
  • @dbaupp: I missed a sign, thanks for pointing that out! – Gigili Jun 08 '12 at 13:23
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    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 08 '12 at 13:51
  • @Gigili: should we consider 4 cases while solving the last absolute value equality?Case 1.both the second and third absolute values are positive 2.both are negative 3.second is positive and third is negative 4.second is negative and third is positive. – mgh Jun 08 '12 at 14:16
  • No, @meg_1997. In this case, there's no variable and you know the term should be positive. Considering the first term, it's positive so it's equal to $\sqrt 5 + \sqrt 7$. The second should be positive, therefore $\sqrt 7 - \sqrt 3$ and so on. – Gigili Jun 08 '12 at 14:26
  • @Gigili: ya got it..thanks – mgh Jun 08 '12 at 14:30
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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: $\rm\ \sqrt{b}+\sqrt{a}.$

See this answer for general radical denesting algorithms.

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Bill Dubuque
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