<s>[INST] How to get the value of the root? I have this statement:

If $\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3}$, Which of
  the following values are the closest to $\sqrt{21}$ ?
A) 68/15 B) 14/3 C) 19/4 D) 55/12 E) 9/2

My development was:
$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}}  = \frac{\sqrt{35}-\sqrt{21}}{2} \approx \frac{2}{3}$
My idea was to treat the sign $\approx$ as a sign $=$ and thus eliminate roots and clear $\sqrt21$, with which I have obtained $55/12$ but I do not know if this is correct. 
$\frac{\sqrt{35}-\sqrt{21}}{2} = \frac{2}{3}$
$35 = (\frac{4}{3} +\sqrt{21})^2$
$\sqrt{21} = \frac{110}{9} * \frac{3}{8} = 55/12$
So my doubt is: Can I treat a $\approx$ sign as a $=$ sign to work like an normal equation?
 [/INST] Rearrange:
$$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3} \iff \frac{7}{\sqrt{35}+\sqrt{21}} \approx \frac{2}{3} \Rightarrow \sqrt{21}\approx \frac{21}{2}-\sqrt{35}$$
Use linear approximation to find $f(35)=\sqrt{35}$:
$$f(x_0+\Delta x)\approx f(x_0)+f'(x_0)\cdot \Delta x \ ;\\
f(36-1)\approx f(36)+\frac1{2\sqrt{36}}\cdot (-1)=6-\frac1{12}=\frac{71}{12}$$
So:
$$\frac{21}{2}-\frac{71}{12}=\frac{126-71}{12}=\frac{55}{12}.$$
</s>