0
$\begingroup$

This expression can be simplified as: $$\sqrt{(x-\frac32)^2} = x - \frac32$$

Even though: $$k^2 = m^2 + n^2 \to \sqrt{k^2} = \sqrt{m^2 + n^2} \to k = \pm\sqrt{m^2 + n^2}$$

You can not remove the right-hand radical sign in the second example even though it is allowed in the first example.

Why is this so? Why can this be applied to other problems?

$\endgroup$
  • 2
    $\begingroup$ In general, $\;\sqrt[n]{x^n}=|x|\;$ for even $\;n\;$ $\endgroup$ – DonAntonio May 1 '16 at 20:01
1
$\begingroup$

Square root is a well defined function, in other words, for any element in the domain there is only one unique function value.

When you solve $x^2=9$, you square root both side, that gives you $\sqrt{x^2}=3$ (3 here is the unique value give by the square root).

Then $ \sqrt{x^2}= |x|=3$ , which gives you two possible solution. $x=3$ or $x=-3$

The two solution is due to the definition of an absolute value, not by the property of square root function.

$\endgroup$
  • 1
    $\begingroup$ Well... I'd like to think that we make the square root function well-defined by choosing for it to always return the principal square root (because square roots are not, in general, unique, although the principal square root is). $\endgroup$ – pjs36 May 1 '16 at 20:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.