Question

Simplify: $\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}$

• A. $2\sqrt{6}$
• B. $1$
• C. $2$
• D. $14$

Answer 1

Answer: (C)

$2$

Squaring the given expression we get:

${(\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}})}^2={\left(\sqrt{7+2\sqrt{6}}\right)}^2+{\left(\sqrt{7-2\sqrt{6}}\right)}^2-(2\times \left(\sqrt{7+2\sqrt{6}}\right)\times \left(\sqrt{7-2\sqrt{6}}\right))\{\because {(a-b)}^2=({a^2+b}^2-2ab)\}=7+2\sqrt{6}+7-2\sqrt{6}-(2\times \left(\sqrt{(7+2\sqrt{6})(7-2\sqrt{6})}\right))=14-(2\times \left(\sqrt{{\left(7\right)}^2-{\left(2\sqrt{6}\right)}^2}\right))\{\because (a-b)(a+b)=a^2-b^2\}$

$=14-(2\times \left(\sqrt{49-(4\times 6)}\right))=14-(2\times \left(\sqrt{49-24}\right))=14-(2\times \sqrt{25})=14-(2\times 5)=14-10=4$

Now if ${(\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}})}^2=4$, then we have:

${(\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}})}^2=4\Rightarrow \sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=\sqrt{4}\Rightarrow \sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=2$

Hence, $\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=2$