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I have the following expression: $a\sqrt{1 + \frac{b^2}{a^2}}$. If I plug this into Wolfram Alpha, it tells me that, if $a, b$ are positive, this equals $\sqrt{a^2 + b^2}$.

How do I get that result? I can't see how that could be done. Thanks

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Hint: for positive $a$, $a = \sqrt{a^2}$. – Mike Wierzbicki Feb 19 '12 at 18:09
up vote 3 down vote accepted

If $a\ge0$, $$\begin{align} a\sqrt{1 + \frac{b^2}{a^2}} &=\sqrt{a^2}\sqrt{1 + \frac{b^2}{a^2}} \\ &=\sqrt{a^2\left(1 + \frac{b^2}{a^2}\right)} \\ &=\sqrt{a^2 + b^2}. \end{align}$$

($\sqrt{a^2}=|a|$ for all $a\in\mathbb{R}$ and $|a|=a$ when $a\ge0$.)

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$$a\sqrt{1 + \frac{b^2}{a^2}}$$

$$=a\sqrt{\frac{a^2 + b^2}{a^2}}$$

$$=a\frac{\sqrt{a^2 + b^2}}{|a|}$$

So when $a$ and $b$ are positive, $|a|=a$. Hence:

$$=\sqrt{a^2 + b^2}$$

Without the assumption:

$$\sqrt{a^2} =|a|=\begin{cases} a && a \geq 0\\ -a &&a < 0\\ \end{cases}$$

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