Question

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

Answer 1

Given expression is $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$

Taking LCM we have:

$$\begin{gathered} \frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}=\frac{4\left(3\sqrt{3}+2\sqrt{2}\right)+3\left(3\sqrt{3}-2\sqrt{2}\right)}{\left(3\sqrt{3}-2\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)} \\ =\frac{12\sqrt{3}+8\sqrt{2}+9\sqrt{3}-6\sqrt{2}}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2}[\left(a+b\right)\left(a-b\right)=a^2-b^2] \\ =\frac{21\sqrt{3}+2\sqrt{2}}{27-8} \\ =\frac{21\left(1.732\right)+2\left(1.414\right)}{19} \\ =\frac{36.372+2.828}{19} \\ =2.063 \end{gathered}$$

So, the value of $\frac{4}{3\sqrt{3}-2\sqrt{2}}+\frac{3}{3\sqrt{3}+2\sqrt{2}}$$=2.063$