Question

Represent the following mixed infinite decimal periodic fractions as common fractions:
${(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})}^2$

Answer 1

Here we can apply $(a+b{)}^2$=$a^2+b^2+2ab$

So here $a=\sqrt{2+\sqrt{3}}$

Now $b=\sqrt{2-\sqrt{3}}$

Now applied it in formula, we have

${\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}^2=$ $(\sqrt{2+\sqrt{3}}{)}^2+$ $(\sqrt{2-\sqrt{3}}{)}^2+$ $2\times \sqrt{2+\sqrt{3}}\times \sqrt{2-\sqrt{3}}$

$=2+\sqrt{3}+$ $2-\sqrt{3}+$ $2\times \sqrt{(2+\sqrt{3})(2-\sqrt{3})}$

$=2+\sqrt{3}+2-\sqrt{3}+2\times \sqrt{2^2-{\sqrt{3}}^2}$

By simplifying it we get

$2+2+2=6$

Hence, the correct answer is $6$.