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Show that : $ \frac{12}{3+\sqrt{5}-2 \sqrt{2}}=\sqrt{10}+\sqrt{5}-\sqrt{2}+1 $
To do:
We have to show that \( \frac{12}{3+\sqrt{5}-2 \sqrt{2}}=\sqrt{10}+\sqrt{5}-\sqrt{2}+1 \).
Solution:
Rationalising the denominator, we get,
$ \begin{array}{l}
\frac{12}{3+\sqrt{5} -2\sqrt{2}} =\frac{12\left( 3+\sqrt{5} +2\sqrt{2}\right)}{\left( 3+\sqrt{5} -2\sqrt{2}\right)\left( 3+\sqrt{5} +2\sqrt{2}\right)}\\
=\frac{36+12\sqrt{5} +24\sqrt{2}}{\left( 3+\sqrt{5}\right)^{2} -\left( 2\sqrt{2}\right)^{2}}\\
=\frac{36+12\sqrt{5} +24\sqrt{2}}{9+5+6\sqrt{5} -8}\\
=\frac{36+12\sqrt{5} +24\sqrt{2}}{6+6\sqrt{5}}\\
=\frac{6+2\sqrt{5} +4\sqrt{2}}{1+\sqrt{5}}\\
=\frac{6+2\sqrt{5} +4\sqrt{2}\left(\sqrt{5} -1\right)}{\left(\sqrt{5} +1\right)\left(\sqrt{5} -1\right)}\\
=\frac{6\sqrt{5} -6+2\sqrt{5\times 5} -2\sqrt{5} +4\sqrt{2\times 5} -4\sqrt{2}}{5-1}\\
=\frac{4\sqrt{5} -6+10+4\sqrt{10} -4\sqrt{2}}{4}\\
=\frac{4\left(\sqrt{10} +\sqrt{5} -\sqrt{2} +1\right)}{4}\\
=\sqrt{10} +\sqrt{5} -\sqrt{2} +1
\end{array}$
Hence proved.
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