How to get the value of the root?

Question

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?

Answer 1

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: $$\frac{21}{2}-\frac{71}{12}=\frac{126-71}{12}=\frac{55}{12}.$$

Answer 2

From $$\left(\frac23\right)^2\approx\left(\frac{\sqrt7}{\sqrt5+\sqrt3}\right)^2$$ we get $$\frac49\approx\frac{7}{5+3+2\sqrt{15}},$$ hence $\sqrt{15}\approx31/8$. Now we have $$\frac{2\sqrt{3}}3\approx \frac{\sqrt7\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt{21}}{\sqrt5+\sqrt3}.$$ Using the approximate value for $\sqrt{15}$ we arrive in $$\sqrt{21}\approx\frac{55}{12}.$$