# Simplify these expressions with radical sign 2

My question is

1) Rationalize the denominator: $$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$

My answer is: $$\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{18}$$

My question is

2) $$\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$$

My answer is: $$\frac{1}{\sqrt{2}}$$

I would also like to know whether my solutions are right. Thank you,

-
in the first 18 should be 12. – S. Snape Jun 8 '12 at 17:23

Multiplying by$$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$ and simplifying will give your that answer:

$$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2 \sqrt 6}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}$$ 2. Your answer is correct.

Multiplying the first fraction by $$\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ And the second by $$\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\sqrt{2}-\sqrt{3}+\sqrt{5}}$$

-

\begin{align} \dfrac1{\sqrt{2} + \sqrt{3} + \sqrt{5}} & = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(\sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3} - \sqrt{5})}\\ & = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(\sqrt{2} + \sqrt{3})^2 - 5}\\ & = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{(2+3+2 \sqrt{6}) - 5}\\ & = \dfrac{\sqrt{2} + \sqrt3 - \sqrt5}{2 \sqrt{6}}\\ & = \dfrac{(\sqrt{2} + \sqrt3 - \sqrt5) \sqrt{6}}{2 \times 6}\\ & = \dfrac{(\sqrt{12} + \sqrt{18} - \sqrt{30} )}{12}\\ \end{align} Hence, your denominator should be $12$ and not $18$ for the first problem.

For the second problem, \begin{align} \dfrac1{\sqrt{2} + \sqrt{3} - \sqrt{5}} + \dfrac1{\sqrt{2} - \sqrt{3} - \sqrt{5}} & = \dfrac1{(\sqrt{2} - \sqrt{5}) + \sqrt{3}} + \dfrac1{(\sqrt{2} - \sqrt{5}) - \sqrt{3}} \\ & = \dfrac{2(\sqrt{2} - \sqrt{5})}{(\sqrt{2} - \sqrt{5})^2 - 3}\\ & = \dfrac{2(\sqrt{2} - \sqrt{5})}{7 - 2 \sqrt{10} - 3}\\ & = \dfrac{\sqrt{2}(2 - \sqrt{10})}{4 - 2 \sqrt{10}}\\ & = \dfrac{\sqrt{2}(2 - \sqrt{10})}{2(2 - \sqrt{10})}\\ & = \dfrac{\sqrt{2}}{2} \end{align} Hence, your second answer is indeed correct. You may want to rationalize the denominator by multiplying by $\sqrt{2}$.

-
@meg: Complete answer. – S. Snape Jun 8 '12 at 17:33
You might want to write $2\sqrt{3}$ instead of $\sqrt{12}$ and $3\sqrt{2}$ instead of $\sqrt{18}$. – Michael Hardy Jun 8 '12 at 18:04

$\begin{eqnarray*} (\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{12}+\sqrt{18}-\sqrt{30}) & = & (\sqrt{2}+\sqrt{3}+\sqrt{5})(2\sqrt{3}+3\sqrt{2}-\sqrt{2}\sqrt{3}\sqrt{5})\\& = & 12, \end{eqnarray*}$

if you expand out the terms, so your first answer is incorrect. The denominator should be $12$.

$\begin{eqnarray*} (\sqrt{2}+\sqrt{3}-\sqrt{5})(\sqrt{2}-\sqrt{3}-\sqrt{5}) & = & (\sqrt{2}-\sqrt{5})^2-\sqrt{3}^2\\& = & 7-2\sqrt{10}-3\\& = & 2\sqrt{2}(\sqrt{2}-\sqrt{5}), \end{eqnarray*}$

and so when your fractions in the second part are given common denominators, you'll have exactly $\cfrac{1}{\sqrt{2}}$ after cancellation, so your second answer is correct.

Note: In general, if you want to see if two fractions are the same (as in the first problem), cross-multiplication is often a useful way to see it.

-