How can someone compute step by step the function $\sqrt{x^2+1}$ to the form $|x|\sqrt{1+ \frac{1}{x^2}}$?

The only way I know how to take something out of a root is to factor inside the root and if that factor can be squared then I would square it and get rid of the power and put it outside the root.

For further clarification of the concept $\sqrt{x^3 + x^2}$ to $\sqrt{x^2(x+1)}$ to ${|x|}\sqrt{x+1}$.

But that is not applicable here because the function $\sqrt{x^2+1}$ cannot be factored, at least a $x^2$ cannot be factored out of $x^2+1$

So how is done?

  • 3
    $\begingroup$ What is $ x^2 \left( 1 + \frac{1}{x^2} \right)? $ $\endgroup$
    – Will Jagy
    May 1 '15 at 4:30
  • $\begingroup$ Yes, the factoring shouldn't be done since it produces new discontinuities, like at $x=0$. Also, there should be a $|x|$ instead of $x$ outside the square root in the factored part considering that we have $x\in\Bbb{R}\setminus\{0\}$ $\endgroup$ May 1 '15 at 4:31
  • $\begingroup$ In general, remember that $\sqrt{xy}=\sqrt{x}\cdot\sqrt{y}~\forall~x,y\geq 0$. $\endgroup$ May 1 '15 at 4:33

$$\sqrt {x^2+1}=\sqrt {x^2*1+x^2*\dfrac {1}{x^2}}$$ as $a*\dfrac {1}{a}=1$. Also, $a*b+a*c=a*(b+c)$, which gives,

$$\sqrt {x^2*1+x^2*\dfrac {1}{x^2}}=\sqrt {x^2*\left (1+\dfrac {1}{x^2}\right)}$$ Now, $\sqrt {ab}=\sqrt {a}*\sqrt {b}$, for non-negative $a$ and $b$. Clearly, $x^2$ and $\dfrac {1}{x^2}$ are non-negative. Therefore,

$$\sqrt {x^2*\left (1+\dfrac {1}{x^2}\right)}=\sqrt {x^2}*\sqrt {1+\dfrac {1}{x^2}}$$ $$=|x|*\sqrt {1+\dfrac {1}{x^2}}$$


The gist is this. As long as $x \gt 0$ the following string of equalities is true.

$$ \sqrt{x^2+1} = \sqrt{x^2 \left({1 + \frac{1}{x^2}}\right)} = \sqrt{x^2} \cdot \sqrt {\left({1 + \frac{1}{x^2}}\right)} = x\sqrt{1+ \frac{1}{x^2}} $$

Don't think of it as a computation. Although if you must the second equality does constitute a "factoring" but the factors are not polynomials.


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