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Could someone please help me solve this problem? I want to be able to determine if in the following separate problems the real numbers can be expressed as indicated $$2\sqrt{2}=\frac{\sqrt{a+b\sqrt{c}}}{d}\pm\frac{\sqrt{a-b\sqrt{c}}}{d}$$ and $$7=\frac{\sqrt{a+b\sqrt{c}}}{d}\pm\frac{\sqrt{a-b\sqrt{c}}}{d}$$ where $a,b,c,d$ are integers and if so is there just one, or are there more solutions? I have squared the LHS and RHS to see where that would lead me but to no avail. Many thanks.

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By using the $\pm$ sign, you mean that the number can be expressed by either a sum or difference of the two fractions, but not necessarily by both, correct? –  anorton Jan 25 '13 at 19:59
    
If there are no restrictions on $a,b,c,d$ one can find many solutions. –  André Nicolas Jan 25 '13 at 20:00
    
@AndréNicolas: The restriction is where $a,b,c,d$ are integers... –  anorton Jan 25 '13 at 20:03
    
@anorton that is correct the number can either be a sum or a difference of the two fractions or both I guess if there can be more than one solution. –  user59670 Jan 25 '13 at 20:03
    
Yes, that was in the post, but I meant further restrictions. –  André Nicolas Jan 25 '13 at 20:04

5 Answers 5

up vote 1 down vote accepted

Well, if they were NOT separate problems, $$ \frac{\sqrt{57+28\sqrt{2}}}{2}+\frac{\sqrt{57-28\sqrt{2}}}{2} = 7,\qquad \frac{\sqrt{57+28\sqrt{2}}}{2}-\frac{\sqrt{57-28\sqrt{2}}}{2} = 2\sqrt{2} . $$

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Write $(7+2\sqrt{2})^2 = a+b\sqrt{2}$ and then $a,b,c=2,d=2$ is a solution for the first equation with "+" and the second equation with "-".

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I don’t see how $$2\sqrt{2}=\frac{\sqrt{2+2\sqrt{2}}}{2}+\frac{\sqrt{2-2\sqrt{2}}}{2}$$ if that is what you are saying for first equation since the second part of the RHS is a complex number. –  user59670 Jan 25 '13 at 20:54
    
No, $a,b$ are computed in the first step, and are not $2$. –  Thomas Andrews Jan 25 '13 at 22:08

For the first, take for example $a=4$, $b=2$, $c=4$, $d=1$.

Or else take $a=3$, $b=2$, $c=2$, $d=1$.

There are infinitely many other choices. In this case, one could describe them all.

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hint: compute $$2\sqrt{2}+7$$ $$2\sqrt{2}-7$$ $$(2\sqrt{2})\cdot7$$

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2  
\cdot to get $\cdot$ –  Git Gud Jan 25 '13 at 20:02
    
thanks Git Gud ..... –  Maisam Hedyelloo Jan 25 '13 at 20:08

Here is one approach to come up with a set of solutions. $2\sqrt{2} =\sqrt{8}$ and the number $8$ can be calculated an infinite number of ways. For example $7+1$; $6+2$; $5+3$; $4+4$; $3+5$; $2+6$; $1+7$; $9-1$; $10-2$.... So let’s say we want to compute$$ \sqrt{100-92}$$ we can construct the following quadratic equation from this $$x^2-100x+\frac{92^2}{4}=0$$ The solutions to this quadratic are $$50+8\sqrt{6}$$ and $$50-8\sqrt{6}$$ Because we want the $\sqrt{8}$ we take the square roots of each of the solutions for the result$$\sqrt{50+8\sqrt{6}}-\sqrt{50-8\sqrt{6}}=2\sqrt{2}$$ Likewise with $$ \sqrt{7+1}$$ we can construct the following quadratic equation from this $$x^2-7x+\frac{1^2}{4}=0$$ The solutions to this quadratic are $$\frac{7+4\sqrt{3}}{2}$$ and $$\frac{7-4\sqrt{3}}{2}$$ Because we want the $\sqrt{8}$ we take the square roots of each of the solutions for the result$$\frac{\sqrt{14+8\sqrt{3}}}{2}+\frac{\sqrt{14-8\sqrt{3}}}{2}=2\sqrt{2}$$ So it appears as has been stated in other answers there are infinite solutions. The same could be done with the equation $$7=\frac{\sqrt{a+b\sqrt{c}}}{d}\pm\frac{\sqrt{a-b\sqrt{c}}}{d}$$ using combinations that equal $\sqrt{49}$.

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