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

**mathwizard** · June 17th 2008, 11:45 AM

Simplify

$<br /> \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

EDIT: Perfect!

**Mathstud28** · June 17th 2008, 11:59 AM

Originally Posted by mathwizard

Simplify

$<br /> \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

Before I type it all up is the answer

$(\sqrt{3}-1)\cdot\sqrt{5}+\sqrt{3}-2$ ?

**Soroban** · June 17th 2008, 01:04 PM

Hello, mathwizard!

Simplify: . $<br /> \frac{\sqrt{3+\sqrt{5}} + \sqrt{4-\sqrt{15}}} {\sqrt{3+\sqrt{5}} - \sqrt{4-\sqrt{15}}}$

Those expressions made me suspicious . . .

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

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

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

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

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

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

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

We agree, Mathstud28 !!

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

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