[br][color=#0000ff]Let [math]\large \textcolor{blue}{x\geq 0.}[/math] The [i][b]square root of [/b][/i][b]x[/b] means the real number, for which[/color][br] [br]  [math]\large \textcolor{blue}{\sqrt x \geq 0 \;\;\text{ and }\;\; (\sqrt x)^2=x.}[/math][br] [br][br]Squaring can be done separately, if all conditions are satisfied:[br][br]  1. [math]\large \textcolor{blue}{\sqrt{ab}=\sqrt a \sqrt b,\;\;\;a,b \geq 0}[/math][br][br]  2. [math]\large \textcolor{blue}{\sqrt\frac a b=\frac \sqrt a \sqrt b,\;\;\;\;\;\;\; a\geq 0, \; b>0 }[/math][br] [br][br]Examples:[br][br] 1.  [math] \sqrt 4=\sqrt{2^2}=2[/math][br] [br] 2.  [math] \sqrt 4=\sqrt{(-2)^2}=|-2|=2 [/math][br] [br] 3.  [math] \sqrt{16+9}=\sqrt{25}=5[/math][br] [br] 4.  [math] \sqrt{2^2+6^2}=\sqrt{4+36}=\sqrt{4\cdot 10}=\sqrt 4\cdot \sqrt{10}=2\sqrt{10}[/math][br] [br] 5.  [math]\sqrt\frac 4 9=\frac \sqrt 4 \sqrt 9=\frac 2 3 [/math][br] [br][br][br][color=#0000ff]Example 6[/color]. [math] \sqrt{16ab^2}=\sqrt{16}\sqrt a\sqrt {b^2}=4\sqrt a|b| [/math][br][br]In this example, the parameter [i]a[/i] must be postive. However, the same deduction cannot be done for the parameter [i]b, [/i] as squaring loses information on the sign:[br][br][br][color=#0000ff]Example 7. [/color] [math] \sqrt \frac{9a^3b}{b^3c^4}=\sqrt\frac{9a^3}{b^2c^4}=\frac{\sqrt 9\sqrt{a^2\cdot a}}{\sqrt {b^2}\sqrt{c^4}}=\frac{3a\sqrt a}{|b|c^2} [/math][br][br][br][color=#0000ff]Example 8. [/color] [math] \sqrt{ab}\left (\sqrt\frac a b+\sqrt \frac b a\right )=\sqrt{ab}\sqrt\frac a b+\sqrt{ab}\sqrt\frac b a =\sqrt \frac{ab\cdot a} b+\sqrt \frac{ab\cdot b} a =|a|+|b|[/math] [br][br]