Question

After rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$, we get the denominator as __________.

Answer 1

$$\begin{gathered} \frac{7}{3\sqrt{3}-2\sqrt{2}} \\ \mathrm{Multiply}\mathrm{and}\mathrm{divide}\mathrm{by}\left(3\sqrt{3}+2\sqrt{2}\right),\mathrm{we}\mathrm{get} \\ =\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}} \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:\hspace{0.17em}\left(a+b\right)\left(a-b\right)=a^2-b^2\right) \\ =\frac{21\sqrt{3}+14\sqrt{2}}{27-8} \\ =\frac{21\sqrt{3}+14\sqrt{2}}{19} \end{gathered}$$


Hence, after rationalizing the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$, we get the denominator as $\underline{\frac{21\sqrt{3}+14\sqrt{2}}{19}}.$