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These are two questions from a competitive exam involving irrationals where I am supposed to simplify it to match one of the given options.

QUESTION 1: The value of $$ \frac {2 (\sqrt 2+ \sqrt6)}{3(\sqrt {2 + \sqrt 3})} + \sqrt {2 + \sqrt 3}+ \sqrt {2 - \sqrt 3}$$ is

A. $\frac {3+4 \sqrt 6}{3}$

B. $\frac {4+3 \sqrt 6}{3}$

C. $\frac {3+4 \sqrt 6}{4}$

D. $\frac {4- 3\sqrt 6}{3}$

I have been able to solve this far:

Taking denominator common, I get

$$ \frac {2 (\sqrt 2+ \sqrt6) + 3(\sqrt {2 + \sqrt 3})(\sqrt {2 + \sqrt 3})+ 3(\sqrt {2 + \sqrt 3})\sqrt {2 - \sqrt 3}}{3(\sqrt {2 + \sqrt 3})}$$

After which:

$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3(\sqrt {2^2 - \sqrt 3^2})}{3(\sqrt {2 + \sqrt 3})}$$

which gives

$$ \frac {2 (\sqrt 2+ \sqrt6) + 3( {2 + \sqrt 3})+ 3}{3(\sqrt {2 + \sqrt 3})}$$

I multiply by the conjugate of the irrational term in denominator.

$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3( {2 + \sqrt 3})\sqrt {2 - \sqrt 3}+ 3(\sqrt {2 - \sqrt 3})}{3(\sqrt {2 + \sqrt 3})(\sqrt {2 - \sqrt 3})}$$

Upon simplification,

$$ \frac {2 (\sqrt 2+ \sqrt6)(\sqrt {2 - \sqrt 3}) + 3(\sqrt 3)\sqrt {2 - \sqrt 3}+ 9(\sqrt {2 - \sqrt 3})}{3}$$

Beyond this, I am not able to work out a solution. Any hints (please explain the hint slightly) are welcome.

QUESTION 2: The value of $$ \sqrt {43-12 \sqrt 7} - \frac {2}{\sqrt {16+6 \sqrt 7}}$$

is:

A. $-3$

B. $3 $

C. $2 \sqrt 7 -3 $

D. $- (2 \sqrt 7 +3) $

I have been able to solve this far:

Taking denominator common:

$$ \frac {\sqrt {43-12 \sqrt 7}(\sqrt {16+6 \sqrt 7}) - 2}{\sqrt {16+6 \sqrt 7}}$$

Upon simplification I get:

$$ \frac {\sqrt {184 + 66\sqrt 7} - 2}{\sqrt {16+6 \sqrt 7}}$$

Multiplying by the conjugate of the denominator:

$$ \frac {\sqrt {184 + 66\sqrt 7}(\sqrt {16-6 \sqrt 7}) - 2(\sqrt {16-6 \sqrt 7)}}{\sqrt {16+6 \sqrt 7}(\sqrt {16-6 \sqrt 7)}}$$

Simplification gives:

$$ \frac {\sqrt {172 - 48\sqrt 7}- 2(\sqrt {16-6 \sqrt 7)}}{\sqrt 4}$$

When I typed this into a software I got

$\sqrt {16-6 \sqrt 7} = 3 - \sqrt 7 $ and

$ \sqrt {172 - 48\sqrt 7} = 12 - 2 \sqrt7 $

How do we get these?

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    $\begingroup$ If you square the expression $$\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}$$ you will find that this is the square root of 6. $\endgroup$ – Joel Oct 31 '14 at 16:40
  • $\begingroup$ Thankyou very much Joel. Amazing!!! But where did you get this from? I mean is there something to look for in the terms when determining hat such a term could be square root of a much simpler term? Thankyou again. $\endgroup$ – Swapnil Rustagi Oct 31 '14 at 16:49
  • $\begingroup$ And you could post this as an answer. This is half of the answer to the question. $\endgroup$ – Swapnil Rustagi Oct 31 '14 at 16:57
  • $\begingroup$ It's really a knee jerk reaction when I see square roots to square them. In some sense these two should balance each other. You can see exactly how when you do square them. In particular this fits the form $$\sqrt{n+\sqrt{n^2-1}} + \sqrt{n-\sqrt{n^2-1}}$$ but other cases should work out too. $\endgroup$ – Joel Oct 31 '14 at 17:23
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As per the request of the OP I am posting my comment as an answer.

If you square the expression $$\sqrt{2+\sqrt{3}} + \sqrt{2-\sqrt{3}}$$ you will find that this is $\sqrt{6}$. This method works for any expression of the form $$\sqrt{n+\sqrt{n^2-1}}+\sqrt{n-\sqrt{n^2-1}}.$$


In order to obtain a simpler version of rational expressions of the form $x=\sqrt{a+b\sqrt{7}}$ we write $$x^2=a+b\sqrt{7}$$ then $$(x^2-a)^2 = 7b^2$$ and so $x$ satisfies the expression $$0=(x^2-a)^2 - 7b^2=x^4-2ax^2 + (a^2 - 7b^2)$$

Next find a factorization for this polynomial, looking for the so called "minimal polynomial" for $x$. That will give you the simplification you desire.

The minimal polynomial corresponding to $\sqrt{43-12\sqrt{7}}$ is $x^2-12x+29$. This means $x = 6 \pm \sqrt{7}$. We conclude that $x=6-\sqrt{7}$ since $6+\sqrt{7} >7$ whose square is larger than 43.


You can factor this polynomial as follows: $$(x^2-a)^2 = 7b^2$$ gives $$x^2 - a = \pm \sqrt{7}b$$ which in turn gives $$x = \pm \sqrt{a \pm \sqrt{7}b}.$$ Thus the polynomial factors as $$\left(x-\sqrt{a - \sqrt{7}b}\right)\left(x-\sqrt{a + \sqrt{7}b}\right)\left(x+\sqrt{a - \sqrt{7}b}\right)\left(x+\sqrt{a + \sqrt{7}b}\right)$$ Picking clever combinations of these will yield the minimal polynomial.

For instance $$\left(x-\sqrt{a - \sqrt{7}b}\right)\left(x+\sqrt{a + \sqrt{7}b}\right)=x^2+x\left(\sqrt{a + \sqrt{7}b}-\sqrt{a - \sqrt{7}b}\right) + \sqrt{a^2-7b^2}.$$ Further by squaring $\sqrt{a + \sqrt{7}b}-\sqrt{a - \sqrt{7}b}$ we see that this is equal to $\sqrt{2a - 2\sqrt{a^2-7b^2}}$. If you are lucky, these will be integers. If not, try combining them differently multiplying terms with $\sqrt{a-\sqrt{7}b}$ against $\sqrt{a+\sqrt{7}b}$. If no combination yields quadratic terms with integer coeficients, there is no quadratic minimal polynomial and the expression cannot be simplified.

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  • $\begingroup$ The minimal polynomial is a polynomial of integer coefficients that divides the quartic polynomial. This doesn't exist for all $a$ and $b$, and in those cases you cannot write a simpler version of your expression. When there is a quadratic minimal polynomial, you can find an expression that involves just one square root. $\endgroup$ – Joel Oct 31 '14 at 19:04
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    $\begingroup$ Well,the minimal polynomial always exists: the thing is, it sometimes has degree 4 :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 31 '14 at 19:52
  • $\begingroup$ That is true. I was sidestepping that statement. $\endgroup$ – Joel Oct 31 '14 at 19:53
  • $\begingroup$ @MarianoSuare-Alvarez, btw if you happen to know a better way to approach this, I would be interested in seeing it. You seem to be a bit more of an algebraist than I am. I am just making this up as I go. $\endgroup$ – Joel Oct 31 '14 at 19:57
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HINT:

$$2+\sqrt3=\cdots=\frac{(\sqrt3+1)^2}2$$

$$\implies\sqrt{2+\sqrt3}=\frac{\sqrt3+1}{\sqrt2}$$

$$\implies\frac{\sqrt6+\sqrt2}{\sqrt{2+\sqrt3}}=\frac{\sqrt2(\sqrt3+1)}{(\sqrt3+1)/\sqrt2}=2$$

Can you derive $$\sqrt{2-\sqrt3}=\frac{\sqrt3-1}{\sqrt2}$$

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  • $\begingroup$ Did it occur to you by chance that 2+ sqrt(3) = (sqrt(3) + 1)^2 / (2) ? Or is there something specific to look for while solving these questions, that could be helpful in simplification? Thankyou very much, very quick reply. $\endgroup$ – Swapnil Rustagi Oct 31 '14 at 16:51
  • $\begingroup$ @Swapnil, It just caught me $4+2\sqrt3=(\sqrt3)^2+1^2+2\sqrt3\cdot1$ $\endgroup$ – lab bhattacharjee Oct 31 '14 at 16:53
  • $\begingroup$ So, I need to be careful and just keep looking out for things that MIGHT help me? But still thanks again. $\endgroup$ – Swapnil Rustagi Oct 31 '14 at 16:56
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    $\begingroup$ Of course you need to keep looking for things that might help you! $\endgroup$ – Mariano Suárez-Álvarez Oct 31 '14 at 19:52
  • $\begingroup$ @MarianoSuárez-Alvarez I realize that. I was asking can this problem be done like differentiation (i.e. you just see it and use the chain, product, or quotient rules which deal with ALL possible causes) instead of integration (where you are left to guess, which method might be more suitable, which susbstition is to be done where, where the factoring of the expression might lead to a simplification to make it a solvable integral). $\endgroup$ – Swapnil Rustagi Nov 1 '14 at 10:03
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You can even express whole root of x-√y in the form of √a-√b . Eg-whole root of 3-√2 can be written as √2-√1

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