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I have this question to express it in a specific form

Express $\dfrac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$ in the form $a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42}$, for some rationals $a,b,c$ and $d$.

So the solution that I have and am trying to understand is this

enter image description here

So what I don't understand is that why in the world is there a 2sqrt6 here where did it come from?

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I would just solve $(a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42})(\sqrt2+\sqrt3+\sqrt7)=1$. –  Berci Nov 11 '12 at 14:02
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3 Answers

up vote 9 down vote accepted

$(\sqrt 2+\sqrt3)^2=2+2(\sqrt 2)(\sqrt3)+3$

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ah i got it now! its the $(a+b)^2$ formula? –  JackyBoi Nov 11 '12 at 13:55
    
yes it is the (a+b)^2 formula. –  Amr Nov 11 '12 at 13:56
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$(\sqrt{2}+\sqrt{3}+\sqrt{7})(\sqrt{2}+\sqrt{3}-\sqrt{7})$

$=(\sqrt{2}+\sqrt{3})^2-(\sqrt{7})^2$

$=(\sqrt{2})^2+(\sqrt{3})^2+2\sqrt{2}\sqrt{3}-(\sqrt{7})^2$

$=2+3+2\sqrt{6}-7$

$=2\sqrt 6-2$

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Nice so clear.. –  JackyBoi Nov 11 '12 at 14:00
    
$2+3+2\sqrt{6}-7$.... I have plus not minus –  JackyBoi Nov 11 '12 at 14:53
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Just for the record, elaborating on Berci's suggestion, you could also try solving $$(a \sqrt{2} + b \sqrt{3} + c \sqrt{7} + d \sqrt{42})(\sqrt{2} + \sqrt{3} + \sqrt{7}) = 1.$$ Writing out the product, you get $$(2a + 3b + 7c) + \sqrt{6}(a + b + 7d) + \sqrt{14}(a + c + 3d) + \sqrt{21}(b + c + 2d) = 1.$$ With all coefficients rational, this implies that the following matrix equation has to be satisfied: $$\begin{pmatrix} 2 & 3 & 7 & 0 \\ 1 & 1 & 0 & 7 \\ 1 & 0 & 1 & 3 \\ 0 & 1 & 1 & 2 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$ There are several ways to solve such systems, all of them leading to the solution: $$\begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} = \frac{1}{10} \begin{pmatrix} 4 \\ 3 \\ -1 \\ -1 \end{pmatrix}.$$ This does not involve any tricks with getting rid of square roots, although one may argue this way to get to the result takes more time.

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