Why does squaring an expression with 2 subtracting terms work?

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?

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

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 $x=-3$