Simplify Fractions: Problem #69 - July 22, 2013

[Jameson]

Without using a calculator and showing your work, simplify $$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}$$.
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) soroban

Solution (from anemone):

Spoiler

First we notice that we can only rewrite the radicand (the sum of two terms) into the form $$\sqrt{m+k\sqrt{n}}=a+b\sqrt{n}$$.

In another word, we could begin to simplify the last two terms first before we start to evaluate the given expression.

Let $$\sqrt{10+2\sqrt{21}}=a+b\sqrt{21}$$.

Squaring both sides to solve for the values for $$ a$$ and $$b$$ we get:

$$10+2\sqrt{21}=a^2+21b^2+2ab\sqrt{21}$$

By equating the coefficient of the $$\sqrt{21}$$ and also the constant we obtain:

$$a=\sqrt{7}$$, $$b=\frac{1}{\sqrt{7}}$$ and hence $$\sqrt{10+2\sqrt{21}}=\sqrt{7}+\left(\frac{1}{\sqrt{7}}\right)\sqrt{21}=\sqrt{7}+\sqrt{3}$$.

We proceed in a similar fashion to simplify $$\sqrt{8+2\sqrt{15}}$$ and get $$\sqrt{8+2\sqrt{15}}=\sqrt{5}+\sqrt{3}$$.

Therefore, we now have

$$\frac{1}{\sqrt{12-2\sqrt{25}}}-\frac{2}{\sqrt{10+2\sqrt{21}}}-\frac{1}{\sqrt{8+2\sqrt{15}}}$$

$$=\frac{1}{\sqrt{12-2\sqrt{25}}}-\frac{2}{\sqrt{7}+\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}$$

What we could do now is to multiply top and bottom of all of these three terms by their conjugates and this yields

$$=\frac{\sqrt{12+2\sqrt{25}}}{(\sqrt{12-2\sqrt{25}})(\sqrt{12+2\sqrt{25}})}-\frac{2(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}-\frac{(1)(\sqrt{5}-\sqrt{3})}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$$

$$=\frac{\sqrt{12+2\sqrt{25}}}{\sqrt{144-140}}-2\left(\frac{\sqrt{7}-\sqrt{3}}{7-3}\right)-\left(\frac{\sqrt{5}-\sqrt{3}}{5-3}\right)$$

$$=\frac{\sqrt{12+2\sqrt{25}}}{2}-2\left(\frac{\sqrt{7}-\sqrt{3}}{4}\right)-\left(\frac{\sqrt{5}-\sqrt{3}}{2}\right)$$

$$=\frac{\sqrt{12+2\sqrt{25}}}{2}-\left(\frac{\sqrt{7}-\sqrt{3}}{2}\right)-\left(\frac{\sqrt{5}-\sqrt{3}}{2}\right)$$

But we know we could simplify $$\sqrt{12+2\sqrt{25}}$$ further to get $$\sqrt{12+2\sqrt{25}}=\sqrt{7}+\sqrt{5}$$ and so

$$=\frac{\sqrt{7}+\sqrt{5}}{2}-\left(\frac{\sqrt{7}-\sqrt{3}}{2}\right)-\left(\frac{\sqrt{5}-\sqrt{3}}{2}\right)$$

$$=\frac{\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{3}-\sqrt{5}+\sqrt{3}}{2}$$

$$=\frac{2\sqrt{3}}{2}$$

$$=\sqrt{3}$$

Alternate solution (from soroban):

Spoiler

$$\text{Simplify: }\:x \;=\; \frac{1}{\sqrt{12-2\sqrt{35}}} - \frac{2}{\sqrt{10+2\sqrt{21}}} - \frac{1}{\sqrt{8+2\sqrt{15}}}$$

Note that: .$$\begin{Bmatrix}12-2\sqrt{35} &=& (\sqrt{7}-\sqrt{5})^2 \\ 10 + 2\sqrt{21} &=& (\sqrt{7}+\sqrt{3})^2 \\ 8 + 2\sqrt{15} &=& (\sqrt{3}+\sqrt{5})^2 \end{Bmatrix}$$We have:
.. $$x \;=\;\frac{1}{\sqrt{7}-\sqrt{5}} - \frac{2}{\sqrt{7}+\sqrt{3}} - \frac{1}{\sqrt{3}+\sqrt{5}}$$

. .$$x \;=\;\frac{\sqrt{7}+\sqrt{3})(\sqrt{3}+\sqrt{5}) - 2(\sqrt{7}-\sqrt{5})(\sqrt{3}+\sqrt{5}) - (\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{34})}{(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{3})(\sqrt{3}+\sqrt{5})}$$The numerator is:

$$N_x \;=\;(\sqrt{7}+\sqrt{3})(\sqrt{3}+\sqrt{5}) - 2(\sqrt{7}-\sqrt{5})(\sqrt{3}+\sqrt{5}) - (\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{34})$$

. . . $$=\;\sqrt{21} + \sqrt{35} + 3 + \sqrt{15} - 2\sqrt{21} - 2\sqrt{35} + 2\sqrt{15} + 10 - 7 - \sqrt{21} + \sqrt{35} + \sqrt{15}$$

. . . $$=\;6 + 4\sqrt{15} - 2\sqrt{21} \;=\;2\sqrt{3}(\sqrt{3} + 2\sqrt{5} - \sqrt{7})$$The denominator is:

$$D_x \;=\;(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{3})(\sqrt{3}+\sqrt{5})$$

. . . $$=\;(\sqrt{7}-\sqrt{5})(\sqrt{21} + \sqrt{35} + 3 + \sqrt{15})$$

. . . $$=\;7\sqrt{3} + 7\sqrt{5} + 3\sqrt{7} + \sqrt{105} - \sqrt{105} - 5\sqrt{7} - 3\sqrt{5} - 5\sqrt{3}$$

. . . $$=\;2\sqrt{3} + 4\sqrt{5} - 2\sqrt{7} \;=\;2(\sqrt{3} + 2\sqrt{5} - \sqrt{7})$$Therefore: .$$x \;=\;\frac{N_x}{D_x} \;=\;\frac{2\sqrt{3}(\sqrt{3}+2\sqrt{5}-\sqrt{7})}{2(\sqrt{3}+2\sqrt{5} - \sqrt{7})} \;=\; \sqrt{3}$$