How can $\sqrt{x^2}=|x|$ [duplicate]

Solution 1:

This is why you need to pay attention to the hypotheses of a theorem! The exponent laws you've learned are almost surely only for integer exponents or for positive bases: until you have learned multi-valued functions and complex exponentiation, you should always be suspicious whenever you're both using non-integers in exponents and negative numbers in bases in the same problem. (presumably, you'd understand the nuances once you've fully learned how to deal with complex analysis)

Solution 2:

It is true that $\sqrt{x} = x^{\frac{1}{2}}$ and $(\sqrt{x})^2 = (x^{\frac{1}{2}})^2 = x$. This is true for all $x$ in the domain of $\sqrt{x}$, namely $x \in [0, \infty)$.

The absolute value comes from composing the square root and square in the opposite order; that is $\sqrt{x^2} = (x^2)^{\frac{1}{2}}$. For any $x \in \mathbb{R}$, $x^2 \geq 0$ so $x^2$ is in the domain of the square root function. The square root function however returns only non-negative values, so if $x < 0$, $(x^2)^{\frac{1}{2}} \neq x$ (but if $x \geq 0$, $(x^2)^{\frac{1}{2}} = x$). In fact, as you have noted, if $x < 0$ write $x = -a$ for some $a \in (0, \infty)$, then $x^2 = (-a)^2 = a^2$, so $(x^2)^{\frac{1}{2}} = (a^2)^{\frac{1}{2}}$. Now $a > 0$ so we obtain $(a^2)^{\frac{1}{2}} = a$, but $a$ is just the absolute value of $x$, i.e. $a = |x|$.

Solution 3:

The principal square root function $f(x) = √x$ is a function that maps the set of non-negative real numbers onto itself. In geometrical terms, the square root function maps the area of a square to its side length. For all real numbers $x$,