2

For negative values of x, $(x^2)^{0.5}$ is $\pm x$.

For negative values of x, $(x^{0.5})^2$ is undefined.

Why do textbooks insist the following: for negative values of $x$, $\sqrt{x^2} = |x|$?

Surely the correct way to think about this is that for negative values of $x$, $\sqrt{x^2}$ is either undefined or equal to $\pm x$, depending on the order the operations are conducted in?

See here for an example: http://www.jamesbrennan.org/algebra/radicals/simplifying_radical_expressions.htm

Caleb Stanford
  • 45,895
Ahsen Majid
  • 21
  • 2
    You seem to have difficulty grasping the concept of "DEFINITION." By definition, the square root of 25 is 5, regardless of the fact that 25 can be thought of as $(-5)^2$. –  Sep 26 '16 at 18:32
  • This question has been asked often. Here's one instance: http://math.stackexchange.com/questions/1059506/must-square-root-of-e-be-positive – Ethan Bolker Sep 26 '16 at 18:37
  • $x^2=w$ is always positive so. So $\sqrt{x^2} = \sqrt{w}$. It doesn't matter a pig's fart what $x$ originally was. The expression $\sqrt{w}$ doesn't "care" how you got the $w$. Once you've got it, you've got it. So $\sqrt{(-5)^2} = \sqrt{25}$. I don't give a crap that $25 = (-5)^2$. All I care about is "what is $\sqrt{25}$". More later.... – fleablood Sep 26 '16 at 18:54
  • $(x^a)^b = (x^b)^a$ is ONLY true if eithe $x \ge 0$ or if $x < 1$ then $a$ and $b$ are integers or if either is rational it has an odd denominator. If $x < 0$ then $x^b$ where b is not rational are is rational with an even denominator is simply not defined. $\sqrt{x^2} \ne \sqrt{x}^2$ unless $x$ is positive. Also if $x < 0$ then $(x^2)^{.5} = -x$ it never is equal to $x$ and for $x < 0$ we can not say $(x^2)^{.5} = x^{.5})^2$ – fleablood Sep 26 '16 at 19:00
  • "For negative values of x, $(x2)^{0.5} = \pm x$" That isn't true. for negative values of $x$, $(x2)^{0.5} = -x = |x|$. – fleablood Sep 26 '16 at 19:04

1 Answers1

1

When $a$ is positive there are indeed always two solutions to the equation $x^2 = a$. Mathematicians have agreed that only the positive one will be called $\sqrt{a}$.

The textbooks insist on that convention.

In any particular algebra problem you may need to think about both solutions.

Edit (in response to the edited question).

When $a$ is negative, that equation has no solution, so you can't talk about $a^{1/2}$. You can talk about $(a^2)^{1/2}$. Whether that's $|a|$ or $\pm a$ depends on the convention for raising to the power $1/2$, which might mean "find the (positive) square root,$\sqrt{a^2}$" or "find both roots". I'm not sure that convention is firmly established.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • 1
    Important to note: If you are using a text that allows $x^{.5} = \pm \sqrt{x}$ then the expression $b^r; r \in \mathbb Q$ does not have a single value and can not be used to denote a single number. There are text that do this but I believe they are only used for a narrow scope. – fleablood Sep 26 '16 at 19:07