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$$\begin{align} v(p_1, p_2, w) & = \sqrt{\frac w{p_1^2\left(\frac1{p_1}+\frac1{p_2}\right)}} + \sqrt{\frac w{p_2^2\left(\frac1{p_1}+\frac1{p_2}\right)}} \\ & = \sqrt{\frac w{\left(\frac1{p_1}+\frac1{p_2}\right)}} \left(\sqrt{\frac1{p_1^2}}+\sqrt{\frac1{p_2^2}}\right) \\ & = \sqrt{\frac w{p_1}+\frac w{p_2}} \end{align}$$

So as the title says, can anyone explain to me this square root?

(Original scan)

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Step by step:

Rewrite the original expression as follows:

$$\sqrt{\frac{w}{p_{1}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} + \sqrt{\frac{w}{p_{2}^{2}(\frac{1}{p_{1}} + \frac{1}{p_{2}})}} = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right)$$

Factoring, this becomes

$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}}\right) + \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right)$$

Note that

$$\sqrt{\frac{1}{p_{1}^{2}}} = \frac{1}{p_{1}}$$

and

$$\sqrt{\frac{1}{p_{2}^{2}}} = \frac{1}{p_{2}}$$

So

$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\sqrt{\frac{1}{p_{1}^{2}}} + \sqrt{\frac{1}{p_{2}^{2}}}\right) = \sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)$$

This can be rewritten as

$$\sqrt{\frac{w}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right) = \sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}}$$

And finally, we can simplify this to

$$\sqrt{\frac{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)^{2}}{\frac{1}{p_{1}}+\frac{1}{p_{2}}}} = \sqrt{w\left(\frac{1}{p_{1}} + \frac{1}{p_{2}}\right)} = \sqrt{\frac{w}{p_{1}} + \frac{w}{p_{2}}}$$

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  • $\begingroup$ Fantastic, very clear, thanks very much :) $\endgroup$ – Maximilian1988 Oct 23 '13 at 5:52
  • $\begingroup$ My pleasure! Glad I could help :) $\endgroup$ – Alex Wertheim Oct 23 '13 at 5:53
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Substituting $q_i=1/p_i$ we find $$q_1\sqrt{\frac{w}{q_1+q_2}}+q_2\sqrt{\frac{w}{q_1+q_2}}=(q_1+q_2)\sqrt{\frac{w}{q_1+q_2}}=\sqrt{w(q_1+q_2)},$$ as $q_1+q_2=\sqrt{(q_1+q_2)^2}$.

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