[tex]\begin{gathered} \frac{\sqrt{8x^2}}{\sqrt{2x}} \\ \\ =\frac{\sqrt{8}\times\sqrt{x^2}}{\sqrt{2}\times\sqrt{x}} \\ \\ =\frac{\sqrt{4\times2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{2}\times\sqrt{x}} \\ \\ =\frac{2\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{2}\times\sqrt{x}} \\ \\ \text{ Cancel out like terms in the numerator and denominator} \\ \\ =\frac{2\times\sqrt{x}}{1} \\ \\ =2\sqrt{x} \\ \\ \text{ Since all the values of the square root can be 0 or greater, the values of x are:} \\ x\ge0\text{ \lparen OPTION A\rparen} \end{gathered}[/tex]
