How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
 A: $f(x) = \sqrt{5 x^2 + 2x +1} $ is the positive function whose square $f^2$ is $f^2=5 x^2 + 2x +1$.
Hence the square $(\frac f x)^2$ is $(\frac f x)^2 = 5 +\frac 2 x + \frac 1 {x^2}$, which implies that 
$$\Big | \frac{\sqrt{ 5x^2 +2x+1}}{x}\Big|= \sqrt { 5 +\frac 2 x + \frac 1 {x^2}}.$$
A: If $x>0$ then $x=\sqrt{x^2}$. So if $x>0$, then
$$\begin{align*}
\frac{1}{x}\cdot\sqrt{5x^2+2x+1} &= \frac{\sqrt{5x^2+2x+1}}{x} \\
&= \frac{\sqrt{5x^2+2x+1}}{\sqrt{x^2}} \\
&= \sqrt{\frac{5x^2+2x+1}{x^2}} \\
&= \sqrt{\frac{5x^2}{x^2}+\frac{2x}{x^2}+\frac{1}{x^2}} \\
&= \sqrt{5+\frac{2}{x}+\frac{1}{x^2}} \\
\end{align*}$$
But if $x<0$, then $x=-\sqrt{x^2}$. So in this case we would get that
$$\begin{align*}
\frac{1}{x}\cdot\sqrt{5x^2+2x+1} &= \frac{\sqrt{5x^2+2x+1}}{x} \\
&= \frac{\sqrt{5x^2+2x+1}}{-\sqrt{x^2}} \\
&= -\sqrt{\frac{5x^2+2x+1}{x^2}} \\
&= -\sqrt{\frac{5x^2}{x^2}+\frac{2x}{x^2}+\frac{1}{x^2}} \\
&= -\sqrt{5+\frac{2}{x}+\frac{1}{x^2}} \\
\end{align*}$$
A: Just a slight variation to the approaches already posted.
Suppose 
$A = \sqrt{5x^2 + 2x + 1}$
$B = \frac{1}{x}$
If $AB = C$, then ${(AB)}^2 = C^2$ or $A^2B^2 = C^2$
$A^2$ = $5x^2 + 2x + 1$
$B^2$ = $\frac{1}{x^2}$
So $C^2 = 5 + \frac{2}{x} + \frac{1}{x^2}$
And to return to C: $$\sqrt{C^2} = \pm\sqrt{5 + \frac{2}{x} + \frac{1}{x^2}}$$
