How do I solve ${2(\sqrt{2}+\sqrt{6})}\over{3(\sqrt{2+\sqrt{3}})}$

Question

I mean, I have found the solution to this problem(4/3) and have even found the solution on Google but am unable to understand it. Moreover, it does not seem to work for other numbers. If anyone could help me solve it and/or give a formula for it, I would really appreciate it. NOTE: this might seem like a very basic question to all you professionals out there, therefore please explain each step thoroughly. Thanks for reading.

Answer 1

First let's manage the denominator: $$ \sqrt{2+\sqrt{3}}=\sqrt{\frac{3+2\sqrt{3}+1}{2}}= \sqrt{\frac{(\sqrt{3}+1)^2}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}} $$ Thus your fraction is $$ \frac{2}{3}(\sqrt{2}+\sqrt{6})\frac{\sqrt{2}}{\sqrt{3}+1} $$ Now note that $\sqrt{2}+\sqrt{6}=\sqrt{2}(1+\sqrt{3})$.

A different strategy is to compute the square: $$ \frac{4}{9}\frac{2+2\sqrt{12}+6}{2+\sqrt{3}}= \frac{4}{9}\frac{4(2+\sqrt{3})}{2+\sqrt{3}} $$

Answer 2

$$x=\frac{2(\sqrt2+\sqrt6)}{3\sqrt{2+\sqrt3}}\Rightarrow x^2=\frac{4(8+2\sqrt{12})}{9(2+\sqrt3)}=\frac{16(2+\sqrt3)}{9(2+\sqrt3)}$$

Thus $x=\dfrac43$

Answer 3

Let $a$ be the number $$a=\frac{2(\sqrt 2+\sqrt 6)}{3\sqrt{2+\sqrt 3}}$$ Squaring both sides of the equation and noting that $(x+y)^2=x^2+2xy+y^2$ gives

$$a^2=\frac{4(2+2\sqrt {12}+6)}{9(2+\sqrt 3)}$$

Noting that $\sqrt{12}=\sqrt{4\cdot3}=2\sqrt 3$ and evaluating the expression on the right hand side above yields

$$a^2=\frac{32+16\sqrt{3}}{18+9\sqrt 3}=\frac{16(2+\sqrt 3)}{9(2+\sqrt 3)}=\frac{16}{9}$$ Therefore $a=\pm4/3$, and since the original expression is clearly positive, $a=4/3.$

Answer 4

multiplying numerator and denominator by $$\sqrt{2-\sqrt{3}}$$ we get $$\frac{2(\sqrt{2}+\sqrt{6})\sqrt{2-\sqrt{3}}}{3(\sqrt{4-3})}$$

As $$(\sqrt{2}+\sqrt{6})\sqrt{2-\sqrt{3}}=\sqrt{(2-\sqrt{3})(8+2\sqrt{3})}=\sqrt{16-8\sqrt{3}+8\sqrt{3}-12}$$

$$=\frac{2}{3}\cdot 2=\frac{4}{3}$$

Answer 5

Hint: $$\color{gray}{2+\sqrt{3}=1/2+\sqrt 3 +3/2=\frac{1+2\sqrt 3 +3}{2}=\frac{1+2\sqrt 3 +(\sqrt3)^2}{2}=\frac{(\sqrt 3+1)^2}{2}}$$

And:

$$\color{gray}{\frac{2(\sqrt 2+\sqrt 6)}{3\frac{\sqrt{(1+\sqrt 3)^2}}{\sqrt 2}}}$$

Answer 6

$$\frac{2(\sqrt{2}+\sqrt{6})}{3(\sqrt{2+\sqrt{3}})}$$$$$$ Multiplying and dividing by $\sqrt{2}-\sqrt{6}$,$$$$ $$\frac{2(\sqrt{2}+\sqrt{6})}{3(\sqrt{2+\sqrt{3}})}\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}$$$$$$ $$\frac{2\color{red}{(\sqrt{2}+\sqrt{6})(\sqrt{2}-\sqrt{6})}}{3(\sqrt{2+\sqrt{3}})(\sqrt{2}-\sqrt{6})}$$$$$$ As $\bbox[5px,border:2px solid red]{(a+b)(a-b)=a^2-b^2}$$$$$ $$\frac{2\color{red}{(2-6)}}{3(\sqrt{2+\sqrt{3}})(\sqrt{2}-\sqrt{6})}$$$$$$ $$\frac{-8}{3(\sqrt{2+\sqrt{3}})(\sqrt{2}-\sqrt{6})}$$$$$$ Multiplying and dividing by $\sqrt{2}$,$$$$ $$\frac{-8\sqrt{2}}{3×\color{red}{\sqrt{2}(\sqrt{2+\sqrt{3}})}(\sqrt{2}-\sqrt{6})}$$$$$$ $$\frac{-8\sqrt{2}}{3\color{red}{(\sqrt{4+2\sqrt{3}})}(\sqrt{2}-\sqrt{6})}$$ $$$$ $$\frac{-8\sqrt{2}}{3(\sqrt{3+2\sqrt{3}+1})(\sqrt{2}-\sqrt{6})}$$$$$$ $$\frac{-8\sqrt{2}}{3(\sqrt{\color{red}{(\sqrt{3})^2+2×\sqrt{3}×1+(1)^2}})(\sqrt{2}-\sqrt{6})}$$$$$$ As $\bbox[5px,border:2px solid red]{a^2+2ab+b^2=(a+b)^2}$$$$$ $$\frac{-8\sqrt{2}}{3(\sqrt{\color{red}{(1+\sqrt{3})^2}})(\sqrt{2}-\sqrt{6})}$$$$$$ $$\frac{-8\sqrt{2}}{3(1+\sqrt{3})(\sqrt{2}-\sqrt{6})}$$$$$$ $$\frac{-8\sqrt{2}}{3(1+\sqrt{3})(\sqrt{2}-\sqrt{3}\sqrt{2})}$$$$$$ $$\frac{-8\sqrt{2}}{3(1+\sqrt{3})(1-\sqrt{3})\sqrt{2}}$$$$$$ $$\frac{-8}{3\color{red}{(1+\sqrt{3})(1-\sqrt{3})}}$$$$$$ As $\bbox[5px,border:2px solid red]{(a+b)(a-b)=a^2-b^2}$$$$$ $$\frac{-8}{3\color{red}{(1-3)}}$$$$$$ $$\frac{-8}{-6}$$$$$$

$$\frac{4}{3}$$$$$$