1. Sep 17, 2015

### basty

If $\sqrt{x^2} = |x|$, why $\sqrt{16} ≠ |4|$ instead of 4 (please see below image)?

2. Sep 17, 2015

### HomogenousCow

The absolute value of 4 is indeed 4. The absolute value is only important when x is negative.

3. Sep 17, 2015

### basty

I mean if $\sqrt{16} = 4$ why $\sqrt{x^2}$ is not x?

4. Sep 17, 2015

### HomogenousCow

Because when x is negative, $\sqrt{x^2} = -x$

5. Sep 18, 2015

### Staff: Mentor

But is $\sqrt{(-4)^2} = -4$?

6. Sep 18, 2015

### basty

Yes indeed.

Because $\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4$

7. Sep 18, 2015

### Staff: Mentor

Absolutely not! $\sqrt{(-4)^2} = \sqrt{16} = 4 = |-4|$

8. Sep 18, 2015

### basty

$\sqrt{(-4)^2} = (-4)^{\frac{2}{2}} = (-4)^1 = -4$

Isn't from the above shows that $\sqrt{(-4)^2} = -4$?

9. Sep 18, 2015

### Staff: Mentor

No, it doesn't.

The exponent properties you are using apply only to numbers that are nonnegative.

$\sqrt{(-4)^2} = [(-4)^2]^{1/2} = 16^{1/2} = + 4$
The rule that you are misusing says that $(a^m)^n = a^{mn}$, provided that $a \ge 0$.