# Simplify [√(3+√5) + √(4-√15)] / [√(3+√5) - √(4-√15)]

• Jun 17th 2008, 11:45 AM
mathwizard
Simplify [√(3+√5) + √(4-√15)] / [√(3+√5) - √(4-√15)]
Simplify

$\displaystyle \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

EDIT: Perfect!
• Jun 17th 2008, 11:59 AM
Mathstud28
Quote:

Originally Posted by mathwizard
Simplify

$\displaystyle \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

Before I type it all up is the answer

$\displaystyle (\sqrt{3}-1)\cdot\sqrt{5}+\sqrt{3}-2$?
• Jun 17th 2008, 01:04 PM
Soroban
Hello, mathwizard!

Quote:

Simplify: .$\displaystyle \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

Those expressions made me suspicious . . .

Note that: .$\displaystyle (1 + \sqrt{5})^2 \:=\:1 + 2\sqrt{5} + 5 \:=\:6 + 2\sqrt{5} \:=\:2(3 + \sqrt{5})$
. . Hence: .$\displaystyle 3 + \sqrt{5} \:=\:\frac{(1 + \sqrt{5})^2}{2}$

Note that: .$\displaystyle (\sqrt{5}-\sqrt{3})^2 \:=\:5 - 2\sqrt{15} + 3 \:=\:8 - 2\sqrt{15} \:=\:2(4-\sqrt{15})$
. . Hence: .$\displaystyle 4 - \sqrt{15} \:=\:\frac{(\sqrt{5}-\sqrt{3})^2}{2}$

The numerator is: .$\displaystyle \sqrt{\frac{(1 + \sqrt{5})^2}{2}} + \sqrt{\frac{(\sqrt{5}-\sqrt{3})^2}{2}} \; = \;\frac{1+\sqrt{5}}{\sqrt{2}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} \;=\; \frac{1 + 2\sqrt{5} - \sqrt{3}}{\sqrt{2}}$

The denominator is: .$\displaystyle \sqrt{\frac{(1 + \sqrt{5})^2}{2}} - \sqrt{\frac{(\sqrt{5}-\sqrt{3})^2}{2}} \;=\; \frac{1+\sqrt{5}}{\sqrt{2}} - \frac{\sqrt{5} - \sqrt{3}}{\sqrt{2}} \;=\;\frac{1 + \sqrt{3}}{\sqrt{2}}$

The problem becomes: .$\displaystyle \frac{\dfrac{1 +2\sqrt{5} - \sqrt{3}}{\sqrt{2}}} {\dfrac{1 + \sqrt{3}}{\sqrt{2}}} \;=\;\frac{1 + 2\sqrt{5}-\sqrt{3}}{1 + \sqrt{3}}$

Rationalize: .$\displaystyle \frac{1 + 2\sqrt{5}-\sqrt{3}}{1 + \sqrt{3}}\cdot{\color{blue}\frac{1-\sqrt{3}}{1-\sqrt{3}}}$

. . which simplifies to: .$\displaystyle \frac{-2\sqrt{3} + 2\sqrt{5} - 2\sqrt{15} + 4}{-2} \;\;=\;\;\boxed{\sqrt{3} - \sqrt{5} + \sqrt{15} -2}$

We agree, Mathstud28 !!