sqrt(8-sqrt(55))+sqrt(8+sqrt(55))

sqrt(8-sqrt(55))+sqrt(8+sqrt(55))

$$\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \quad | \quad \sqrt{ x^2 } \\ \\
= \sqrt{ \left( \sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}} \right)^2 } \\ \\
= \sqrt{ (8-\sqrt{55}) +(8+ \sqrt{55}) +2*\left( \sqrt{8-\sqrt{55}} \right)* \left(\sqrt{8+\sqrt{55}} \right)} \\ \\
= \sqrt{ 16+2*\left( \sqrt{8-\sqrt{55}} \right) * \left(\sqrt{8+\sqrt{55}} \right) } \\ \\
= \sqrt{ 16+2* \sqrt{8^2- 55 } }\\ \\
= \sqrt{ 16+2* \sqrt{64-55} } \\ \\
= \sqrt{ 16+2* \sqrt{9} } \\ \\
= \sqrt{ 16+2* 3 }\\ \\
= \sqrt{ 22 }\\ \\
= 4.69041576$$

$$\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}}$$

This is a tough one!

First I am going to consider      $$8-\sqrt{55}$$

I want to express this as a perfect square.

$$\\8-\sqrt{55}\\\\
=\frac{5}{2}-\sqrt{55}+\frac{11}{2}\\\\
=\frac{25}{10}-\frac{10\sqrt{55}}{10}+\frac{55}{10}\\\\
=\frac{25-10\sqrt{55}+55}{10}\\\\
=\frac{5^2-10\sqrt{55}+(\sqrt{55})^2}{10}\\\\
=\frac{(5-\sqrt{55})^2}{10}\\\\
But\;\; (5-\sqrt{55})^2=(\sqrt{55}-5)^2\;\;$and I want the positive one, so $\\\\
=\frac{(\sqrt{55}-5)^2}{10}\\\\$$

$$\\Hence\\\\
\sqrt{8-\sqrt{55}}\\\\
=\sqrt{\frac{(\sqrt{55}-5)^2}{10}}\\\\
=\frac{\sqrt{55}-5}{\sqrt{10}}}\\\\
=\frac{\sqrt{10*55}-5\sqrt{10}}{10}\\\\
=\frac{\sqrt{2*5*5*11}-5\sqrt{10}}{10}\\\\
=\frac{5\sqrt{22}-5\sqrt{10}}{10}\\\\
=\frac{\sqrt{22}-\sqrt{10}}{2}\\\\$$

NOW, BY THE SAME LOGIC,

$$\\Hence\\\\
\sqrt{8}+\sqrt{55}=\frac{\sqrt{22}+\sqrt{10}}{2}\\\\
SO\\\\
\sqrt{8-\sqrt{55}}+\sqrt{8+\sqrt{55}}\\\\
=\frac{\sqrt{22}-\sqrt{10}}{2}+\frac{\sqrt{22}+\sqrt{10}}{2}\\\\
=\frac{2\sqrt{22}}{2}\\\\
=\sqrt{22}\\\\$$

Very crafty, Melody....!!!

Also well done, heureka...!!!

Your method is a little more intuitive to me, than Melody's.....

But.......either one gets the job done!!!

Thanks Chris,  

Yes, I will admit, I like Heureka's method better too.