Question

Question 3
If $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$, then find the value of $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$

Answer 1

We have, $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$
$=\frac{4(3\sqrt{3}+2\sqrt{2})+3(3\sqrt{3}-2\sqrt{2})}{(3\sqrt{3}-2\sqrt{2})(3\sqrt{3}+2\sqrt{2})}=\frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{(3\sqrt{3}{)}^2-(2\sqrt{2}{)}^2}$
[Using identity, $(a+b)(a-b)=a^2-b^2]$
$=\frac{21\sqrt{3}+2\sqrt{2}}{27-8}=\frac{21\sqrt{13}+2\sqrt{2}}{19}$
$=\frac{21\times 1.732+2\times 1.414}{19}$ [Put $\sqrt{2}=1.414$ and $\sqrt{3}=1.732]$
$=\frac{36.372+2.828}{19}=\frac{39.2}{19}=2.0631=2.063$