How does this simplification work?
$\frac{\sqrt{x^2 + x}}{x} =\sqrt{1 + 1/x} $
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1$x= \sqrt{x^2}$ – May 06 '14 at 18:23
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4Hmm. What if $x=-2$? – David Mitra May 06 '14 at 18:24
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$$\frac{\sqrt{x^2+x}}{x}=\frac{\sqrt{x^2+x}}{\sqrt{x^2}}=\sqrt{\frac{x^2+x}{x^2}}=\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}=\sqrt{1+\frac{1}{x}}$$
Note: this is valid for $x>0$, as then $x=|x|=\sqrt{x^2}$.
For $x<-1$, the LHS is defined, but negative, while the RHS is positive, so they do not agree. For $-1<x<0$, the LHS is undefined as you're taking the square root of a negative number. For $x=-1$, the two sides agree since they're both zero. For $x=0$, neither side is defined since you're dividing by zero.
vadim123
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To be precise, this is only correct if $x$ is positive. For negative values of $x$, it is not true that $x=\sqrt{x^2}$; rather, we should have $x = sgn (x) \sqrt{x^2}$, where $sgn(x)$ is the "sign" or "signum" function, giving 1 or -1 depending on whether $x$ is positive or negative. – mweiss May 06 '14 at 18:25
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