Given
[tex]\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\text{ \_\_\_\_\_\_\_ }3[/tex]To compare.
Explanation:
It is given that,
[tex]\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}\text{ \_\_\_\_\_\_\_ }3[/tex]Now, consider
[tex]\sqrt{p}=\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}[/tex]Squaring both sides implies,
[tex]\begin{gathered} p=(\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}})^2 \\ =(6+\sqrt{11})-2(\sqrt{6+\sqrt{11}})(\sqrt{6-\sqrt{11}})+(6-\sqrt{11}) \\ =12-2(\sqrt{(6+\sqrt{11})(6-\sqrt{11})}) \\ =12-2(\sqrt{36-11}) \\ =12-2(\sqrt{25}) \\ =12-2\times5 \\ =12-10 \\ =2 \end{gathered}[/tex]That impies,
[tex]\sqrt{6+\sqrt{11}}-\sqrt{6-\sqrt{11}}=\sqrt{2}<3[/tex]
